C h a p t e r
5:
Point Scoring
P r o b a b i l i t y
L e s s P i i 5 . 1 - EKpU-rhui
P r o l w b i l i l y . p a t j c i :5f}3
1 . e . g . t:<-<
. f^-: m - ;
turn.
2. a ) O u t c o m e l.-f.i,--
i< ^ ! i iv-i , : .i>..i • • i . , . - a c h
,
.
1
:
1
•T.y.r
j
1 li
1
K
' 'dyer I
p
[ z z T i i z n z i ]
A
T
H
1 HH
' ^ "
[HT
i>
'
'iaver 2
I
'f.ijabili!.- >\
I [
i / e r 1 w i n n i n g is n c -
4 3 , 7 5 % . T h e p r o b a b i l i t y o f P l a y e r 2 winr.
P(Matt win
r.-,.;;, {-i.>
, o r 5 0 % . T h i s is n o t a fair g a m e . P l a y e r 2 h a s
P{Pat wins) =
t
1
more opportunities to win.
•V •^' C h a n c e o f w i n n i n g , s o t h e
i, !!
r
H
M
L e s s o n 5.2: Probability a n d O d d s , page 310
1. a) T h e odds against Marcia passing the dnver's
t e s t o n t h e first try a r e 3 : 5.
iT
e"
I
. . ..©w
8
P(Patwins;
P(Matt win
3r
b | L e t A r e p r e s e n t M a r c i a p a s s i n g h e r d r i v e r s test o n
t h e first t r y .
P(Treena wins) = |
8
P(Leen.;
T h e p r o b a b i i i t y M a r c i a will p a s s h e r d r i v e r ' s test o n
T h i s g a m e is n o t fair, G i n a h a s a 6 in 8 c h a n c e o f
winning.
t h e first t r y is ^ , o r 0 , 6 2 5 .
8
c) Outcome Tabie;
Die 1
I
•
'
1 " •
I
3
4
^
2 . a | T h e p r o b a b i l i t y t h e c o i n is a l o o n i e is ^
, or 0 3
10
5
^ 4 "
10 _
:
6
:
b ) T h e p r o b a b i l i t y t h e c o i n is n o t a l o o n i e is ^ , o r 0 7
10
S o , t h e o d d s a g a i n s t t h e c o i n b e i n g a l o o n i e a r e 7 ; 3,
3. a ) L e t R r e p r e s e n t a r e d c a r d b e i n g d r a w n .
P{P)=26
7
11
^ '
PiR)
P(Annwins) = ^
36
P(Danwins) =
^
36
P(Annwins) = |
P(Danwins) =
|
Each player has an equal chance of winning, so the
g a m e is fair.
3 . N o , e . g , , A c e r t a i n c h a n c e is 1 0 0 % . 1 2 0 % > 1 0 0 %
4. Sample Space
Product
1
3
4
1
1
2
3
4
2
2
4
6
8
3
3
4
^4
6
Sum
1
2 "
1
2
3
]
3
4
4""
'
5
b
2
3
4
9
12
12""
16
F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l u t i o n s M a n u a t
52
=
1
P ( P ) = 0.5
T h e p r o b a b i l i t y t h a t Lily d r a w s a r e d c a r d is 0 . 5 ,
b ) L e t P ' r e p r e s e n t a c a r d t h a t is n o t r e d b e i n g d r a w n .
52
P{e)=26
^ '
52
S o , t h e o d d s in f a v o u r o f t h e c a r d b e i n g r e d a r e
26:26, or 1 : 1 .
c ) Let S r e p r e s e n t a s p a d e b e i n g d r a w n .
P(S') =
PiS) =
l
P(S') =
^
|
The odds against the card being a spade are 3 ; 1
5-1
ci|;
1/
- , f
a face card being drawn.
9. If K a t h e n n e h a d s c o r e d 4 t i m e s o u t of 2 0 . t h e n s h e
h a d not s c o r e d 16 t i m e s o u t o f 2 0 . T h e r e f o r e , t h e
o d d s in f a v o u r a r e 4 : 1 6 , o r 1:4 T h i s c a n b e w r i t t e n a s
1 t o 4 . T h e p r o b a b i l i t y o f h e r s c o n n g is 4 in 2 0 , o r
I : 5. T h e r e f o r e . K a t h e r i n e is n o t c o r r e c t .
1 i
10. a) If J a s o n h a s s c o r e d 5 t i m e s in his last
T h e p r o b a b i l i t y t h e c a r d d r a w n is a f a c e c a r d is -^^ ,
or about 0.231
4. a) L e t A r e p r e s e n t a p p l e j u i c e b e i n g a v a i l a b l e at t h e
grocery store.
10 p e n a l t y s h o t s , t h e n h e d i d n ' t s c o r e 5 t i m e s .
T h e r e f o r e , t h e o d d s in f a v o u r o f s c o n n g a r e 5 ; 5,
or 1 : 1.
b) G i l l e s b l o c k e d 8 of t h e last 10 p e n a l t y s h o t s , a n d
let 2 s h o t s i n . T h e r e f o r e , t h e o d d s in f a v o u r o f J a s o n
s c o n n g o n G i l l e s a r e 2 : 8, or 1 : 4 .
c ) e . g . . J a s o n ' s d a t a r e f l e c t s his r e c o r d a g a i n s t all
g o a l i e s , n o t j u s t G i l l e s . G i l l e s ' d a t a s u g g e s t s t h a t h e is
b c t ' ^ r t h a n a v e r a g e at b l o c k i n g p e n a l t y s h o t s
II
T h e o d d s in f a v o u r o f a p p l e j u i c e b e i n g a v a i l a b l e a r e
I , j i S r e p r e s e n t a p e r s o n b e t w e e n t h e a g e s of 18
u f i d J5 w h o u s e s s o c i a l n e t w o r k i n g .
2 ; 3.
b) T h e o d d s a g a i n s t a p p l e j u i c e b e i n g a v a i l a b l e a r e
3 : 2.
5. Let S r e p r e s e n t t w o s t u d e n t s in M a n o ' s c l a s s
^ '
50
s h a n n g a birthday.
P { S ) = 0.62
P(S)-
T h e p r o b a b i l i t y t h a t a p e r s o n b e t w e e n t h e a g e s o f 18
a n d 3 5 u s e s a s o c i a l n e t w o r k i n g site is 0 . 6 2 .
p ( s ) 4
1 2 . Let IV r e p r e s e n t a w i n . let L r e p r e s e n t a l o s s , a n d
P{S)
let T r e p r e s e n t a t i e .
0.7
T h e p r o b a b i l i t y t w o s t u d e n t s s h a r e a b i r t h d a y is 0 . 7 .
P{W)
3 I I
6. Let C r e p r e s e n t J a m i a c l i m b i n g t o t h e t o p ,
a)
P[C)
P{W)
3
p ( n ^ ^
P i T ) - - ^ ,
p(n^-l
P{T)
1
24
T h e p r o b a b i l i t y of a w i n is 3 in 5 ( 6 0 % ) , t h e p r o b a b i l i t y
of a l o s s is 1 in 5 ( 2 0 % ) , a n d t h e p r o b a b i l i t y of a tie is
1 in 5 ( 2 0 % ) . T h e p r o b a b i l i t i e s a d d u p t o 1 0 0 % .
1
T h e p r o b a b i i i t y J a m i a w i l l m a k e it t o t h e t o p is - , or 0 5.
24 - 1 2
b) P ( C ' )
24
12
P{C')
24
P(C') = |
13. a ) 6 5 % o f t h e a u d i e n c e a r e m a l e , a n d 3 5 % of t h e
a u d i e n c e a r e n o t m a l e . T h e r e f o r e , t h e o d d s in f a v o u r
of s o m e o n e w a t c h i n g Show Trial b e i n g m a l e a r e
6 5 : 3 5 , or 13 : 7.
b) 4 0 % o f t h e a u d i e n c e a r e o v e r 4 5 , a n d 6 0 % of t h e
a u d i e n c e a r e u n d e r 4 5 . T h e r e f o r e , t h e o d d s in f a v o u r
of s o m e o n e w a t c h i n g Show Tnal b e i n g o v e r 4 5 a r e
4 0 : 6 0 , o r 2 : 3.
T h e o d d s a g a i n s t J a m i a m a k i n g it t o t h e t o p a r e
12 : 12, or 1 : 1 .
14. a ) 7 0 % o f t h e p e o p l e v a c c i n a t e d d i d not g e t sick,
c ) e . g . . T h e o d d s a g a i n s t a n d t h e o d d s in f a v o u r a r e
s o 3 0 % of t h e p e o p l e v a c c i n a t e d d i d g e t s i c k . S o , t h e
both 1 ; 1.
o d d s a g a i n s t g e t t i n g s i c k if y o u a r e v a c c i n a t e d a r e
7. If t h e p r o b a b i l i t y of s n o w is 6 0 % , t h e n t h e
p r o b a b i l i t y t h a t of n o s n o w is 4 0 % . T h e r e f o r e , t h e
o d d s a g a i n s t s n o w a r e 4 0 : 6 0 , or 2 : 3.
7 0 : 3 0 . o r 7 ; 3.
b) 4 2 % of t h e p e o p l e w h o w e r e n o t v a c c i n a t e d g o t
sick, s o 5 8 % o f t h e p e o p l e w h o w e r e n o t v a c c i n a t e d
d i d not g e t s i c k . T h e r e f o r e , t h e o d d s a g a i n s t g e t t i n g
sick if y o u a r e not v a c c i n a t e d a r e 5 8 : 4 2 . or 2 9 : 2 1 .
c ) T h e first ratio, 7 : 3, c a n b e w n t t e n a s 4 9 : 21 by
m u l t i p l y i n g b o t h n u m b e r s by 7. S o t h e o d d s of g e t t i n g
sick are 49 : 21 a n d the o d d s against getting sick are
29 : 2 1 .
8. If A l l a n h a s a n 8 % c h a n c e o f h a v i n g r e d - g r e e n
c o l o u r b l i n d n e s s , t h e n h e h a s a 9 2 % c h a n c e of not
having red-green colourblindness. Therefore, the
o d d s in f a v o u r of A l l a n h a v i n g r e d - g r e e n
c o l o u r b l i n d n e s s a r e 8 : 9 2 . or 2 ; 2 3
5-2
C h a p t e r 5: P r o b a b i l i t y
d ) e . g . , Y e s , b e c a u s e y o u r c h a n c e s of n o t g e t t i n g sick
a r e m u c h better.
1 5 . a ) T h e o d d s in f a v o u r o f g e t t i n g 7 p o i n t s , by
s c o r i n g a t o u c h d o w n , a n d t h e c o n v e r t , a r e 5 : 7. T h e
o d d s in f a v o u r of g e t t i n g 3 p o i n t s b y k i c k i n g a field
g o a l , a r e 5 : 1 or 3 5 ; 7.
b ) e . g . . T h e o d d s in f a v o u r o f g e t t i n g 3 p o i n t s a r e
m u c h better, s o t h e c o a c h s h o u l d c h o o s e t h e field
goal.
16. a) T h e o d d s
•
• • •
4 5 : 'no .;..
.
if.
.. n • i • h e
o d d s in f a v o u r e!
.-.i •• i-<; . . h ' c - < h-<'•/;
T h i s is e q u a l to c , >, . o: .
^ <-^-id . c . ,.ur
of Bill w i n n i n g a
- •''
'.• - • |.. i •.
2 0 ; 8 0 , or 1 : 4 .
b | If Bill's 2 0 % s • ! : • . • > - . , >
v-\,- i.,/. .|f ; - . . . p p o r t
will n o w b e 5 5 1 . - • r '
/• ..
^ i. r.. •
w i n n i n g will b e t •'
> f> i j .i-. . r ; .h
w i n n i n g . S o , the
.
11:9,
1 7 . a | e , g , , T h e o d d s in f a v o u r o f p a s s i n g a r e 11 ; 9,
s o t h e p r o b a b i l i t y o f p a s s i n g is slightly m o r e t h a n
50%.
F o r e v e r y 2 0 p e o p l e w h o t a k e t h e e x a m . 11 p a s s .
1 i . e . g . , I p r e f e r u s i n g p r o b a b i l i t y b e c a u s e if I e x p r e s s
t h e p r o b a b i l i t y a s a p e r c e n t , it tells m e h o w m a n y
t i m e s o u t o f a h u n d r e d I c o u l d e x p e c t t h e e v e n t to
o c c u r , e . g . , I p r e f e r u s i n g o d d s b e c a u s e it c o m p a r e s
t h e c h a n c e s for a n d a g a i n s t t h e e v e n t o c c u r r i n g ,
2 0 . a ) L e t A r e p r e s e n t a child b e t w e e n t h e a g e s o f 6
a n d 18 w h o will n e e d c o r r e c t i v e l e n s e s . Let R
r e p r e s e n t a girl b e t w e e n t h e a g e s o f 6 a n d 18 w h o will
need corrective lenses.
141
P(4) = 25.4%
25.4
PiA)
'
PiR)
P(R):
100
'
ICl
li)'.}
141
241
500
127
F'I A
141
500'241
P{Ar
17 9 0 7
R)
120 5 0 0
T h e p r o b a b i i i t y t h a t a r a n d o m l y s e l e c t e d girl b e t w e e n
t h e a g e s o f 6 a n d 18 will n e e d c o r r e c t i v e l e n s e s is
b) L e t A r e p r e s e n t a c h i l d b e t w e e n t h e a g e s o f 6 a n d
18 w h o will n e e d c o r r e c t i v e l e n s e s . Let B r e p r e s e n t a
b o y b e t w e e n t h e a g e s of 6 a n d 18 w h o will n e e d
corrective lenses.
. 1 1
Ptpass = —
F ( p a s s ) = 0,55
P(pass)=55%
T h e p r a c t i c e e x a m s c o s t $ 6 5 . T h e c o s t to r e w n t e t h e
e x a m is $ 2 3 5 .
Y e s . If h e p a y s t h e $ 6 5 . h e c a n r e d u c e t h e 4 5 %
c h a n c e o f h a v i n g to p a y t h e a d d i t i o n a l $ 2 3 5 .
b) N o . e . g . , W i t h o d d s of 17 ; 4 , G r a n t h a s a b o u t a
1 9 % c h a n c e o f failing e v e n w i t h o u t t h e p r a c t i c e
e x a m s , so he should probably not buy t h e m .
Y e s . e.g , W i t h o d d s of 3 : 7. his c h a n c e of failing is
7 0 % w i t h o u t t h e p r a c t i c e e x a m s , s o he s h o u l d b u y
F ( ^ ) = 25.4%
141 * 100
P(4)=2M
^ ^
100
^ '
DiM
p,/1
^
P{B):
100
241
500
OV
.)-
(127
100
—
241
Msoo
them.
18. a ) e.g., If t h e o d d s for a n e v e n t a r e m ; n, t h e n
PiA):
m
and P(4') =
m+ n
PU'):PU) =
m + n
^
m+n
in
m+n
, so
1205
T h e probability that a randomly selected 18-year-old
T h i s ratio is e q u a l to n : m.
b) e g . . T h e p r o b a b i l i t y of t h e e v e n t h a p p e n i n g is
. If t h e o d d s in f a v o u r of rain t o m o r r o w a r e 2 : 3,
127
. So, the odds
1205
in f a v o u r o f a r a n d o m l y s e l e c t e d 1 8 - y e a r - o l d b o y will
n e e d c o r r e c t i v e l e n s e s a r e 127 : ( 1 2 0 5
1 2 7 ) . T h i s is
e q u a l to 127 : 1 0 7 8 .
b o y will n e e d c o r r e c t i v e l e n s e s is
a+ b
t h e n t h e p r o b a b i l i t y is
, or 4 0 % .
2+3
5
c) e.g., The odds against the event happening are
c - a : a. If t h e p r o b a b i l i t y o f w i n n i n g t h e lottery is
the odds against are 999 9 9 9 : 1 .
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5-3
L e s s o n 5.3: P r o b a b i l i t i e s U s i n g C o u n t i n g
M e t h o d s , page 321
52!
niO)
1. Let A r e p r e s e n t a p a s s c o d e w i t h 4 d i f f e r e n t e v e n
d i g i t s , a n d let O r e p r e s e n t all 4 - d i g i t p a s s c o d e s .
T h e r e a r e 5 d i f f e r e n t p o s s i b i l i t i e s for e v e n d i g i t s : 0,
2 , 4 . 6 . a n d 8. S i n c e t h e s e n u m b e r s c a n n o t b e
r e p e a t e d , t h e total n u m b e r o f p a s s c o d e s u s i n g
4 d i f f e r e n t e v e n d i g i t s is e q u a l to 5P4.
(52 - 8 ) ! • 8 !
52!
n{0)
4 4 ! • 8!
h / :,1 :>() I'i 4.-; d r
n(r))
If
M
n(0,,
4
:>
\
•[ 3 J i 2
: ,
,-'C 4./ •>'!
I -J h- 17
; t;
T,
niA)
PiA)
niO)
••(/O - 12u
I digit p a s s c o d e s is 1 0 ^ b e c a u s e
1 h i ' nu'fibt
t h e r e a i c l u posoible d i g i t s t o u s e in 4 s p a c e s , a n d
r e p e a t i n g is a l l o w e d .
PiA)
) 13 17
Pi
niO)
/ !. / '
P^A,
n 431 030
T h e p i o b . j o i l i t y t h a t a h a n d will c o n s i s t o f 8 h e a r t s is
10"
-1
120
P{A)
10 0 0 0
T h e p r o b a b i l i t y t h a t S u n ' s p a s s c o d e is m a d e u p of
f o u r d i f f e r e n t e v e n digits is
120
, or 0 . 0 1 2 or 1.2%.
10 0 0 0
2. Let A r e p r e s e n t a n 8 - c a r d h a n d c o n t a i n i n g 8 h e a r t s ,
a n d let O r e p r e s e n t all 8 - c a r d h a n d s .
In a s t a n d a r d d e c k of c a r d s u s e d for C r a z y E i g h t s ,
t h e r e a r e 13 h e a r t s . T h e r e f o r e , t h e total n u m b e r o f
h a n d s w i t h 8 h e a r t s is laCs.
or a b o u t O OOO 0 0 1 7 1 .
6 431'1''()
3. L e t A r e p r e s e n t B e n a n d J e n b e i n g c h o s e n a s
p r e s i d e n t a n d s e c r e t a r y , a n d let O r e p r e s e n t all
possible committees
S i n c e o r d e r is i m p o r t a n t , t h e n u m b e r of w a y s in w h i c h B e n
a n d J e n c a n b e c h o s e n for p r e s i d e n t a n d s e c r e t a r y is 2P2.
niA)
- ,P,
^ '
(2-2)!
n(A)
13!
n{A)
i
il
120
P{A)
1,0
11
n{A)
0'
(13-8)!
n(A)
13!
n{A)
^
5! • 8!
n(A)-'?
13 I 2 J 1 10 0 81
T h e total n u m b e r o f p o s s i b l e c o m m i t t e e s is 12P2,
b e c a u s e t h e r e a r e 12 p e o p l e t o c h o o s e f r o m , a n d
o r d e r is i m p o r t a n t .
54
^ 2 1 Si
13 12^ 11 J O
9
niA)~^
niO)
n(A) = ( i | g > 3 , ( . ) ( .
niO) =
n(/\)
niO)
13 11 9
T h e total n u m b e r o f h a n d s c o n t a i n i n g 8 c a r d s is
52C8, b e c a u s e t h e r e a r e 5 2 c a r d s t o c h o o s e f r o m , a n d
r e p e a t i n g is n o t p o s s i b l e .
5-4
2 !
io 45
N o w determine the probability.
f . A.
riA)
4
• 8 h ' • / I f . '
nil)',
c-'/o
'jO 4 i .in, 1/
:\ ! c
n(nt
46-45•44!
I i] '.. 1
. A
12!
(12-2)!
12!
10!
niO)
1 2 - 1 1 10!
10!
n(0)
12 11
r?(0)
132
C h a p t e r 5: P r o b a b i l i t y
N o w determine the probabiiity.
b) Let C r e p r e s e n t t h a t t h e r e is a n e q u a l n u m b e r of
b o y s a n d girls o n t h e trip.
niO)
5!
F
6!
(5^2)
>)!2!
6!
5!
P(
,•
1
T h e p r o b a b i l i t y B e n a n d J e n will b e c h o s e n is ^ , or
66
a b o u t 0 . 0 1 5 2 o r 1,52%,
4. a) L
/ . f -resent t h e p o s s i b i l i t y t h a t o n l y b o y s will
b e o n t h e trip, a n d let O r e p r e s e n t all p o s s i b i l i t i e s .
. i
/ I ; '
I
,,{( i
!
>,{>
n(C) = 150
[•
T h e total n u m b e r of w a y s there can be an equal
i=c!
n u m b e r o f b o y s a n d giris o n t h e trip is 1 5 0 .
PiC)
1! - 4 !
n(B):
5-4j
'
niO)
niO)
330
P{C)
=
•
^
11
T h e p r o b a b i l i t y t h a t a n e q u a l n u m b e r of b o y s a n d giris
will g o o n t h e trip is
11!
(11^4)!4!
i b o u t 0.455 or 4 5 , 5 % .
c ) Let D r e p r e s e n t t h a t t h e r e a r e m o r e g i d s t h a n b o y s
o n t h e trip. T h i s is t r u e if 4 giris a n d n o b o y s a r e
s e l e c t e d a n d if 3 giris a n d 1 b o y a r e s e l e c t e d .
11!
7! • 4 !
niO)
•
5
T h e n u m b e r of p o s s i b i l i t i e s of t h e r e b e i n g all b o y s o n
t h e trip is 5.
niO)
n(C)
nio)
1 • 4!
niB)
=
1T10-9-8-7!
7! • 4 !
6!
n(D)
11-10-9-8
4-3-2
niD):
2!4!'
11-10-3
o ( 0 ) - 330
n{D)
T h e total n u m b e r of p o s s i b i l i t i e s f o r s e l e c t i n g f o u r
s t u d e n t s for t h e trip is 3 3 0 .
^.'jt'
5!
^
~ '
1!
r r
o S 4 1
6 5 4'
2!4!
n(D)
6!
'(6-4)141 ( 6 - 3 ) ! i
6!
6!
5!
^'
5,41
"IT
15 i 20 5
n ( D ) = 115
T h e r e a r e 1 1 5 w a y s in w h i c h t h e r e c a n be m o r e giris
t h a n b o y s o n t h e trip,
PiD
n(D)
nio)
P(B}
66
PiD)
115
T h e r e f o r e , t h e p r o b a b i l i t y t h a t o n l y b o y s will b e o n t h e
330
trip IS - L , or 0 . 0 1 5 2 or 1.52%.
66
bb
T h e p r o b a b i l i t y t h a t m o r e giris t h a n b o y s will g o is
^
66
or a b o u t 0 , 3 4 8 o r 3 4 . 8 % .
F o u n d a t i o n s of M a t h e m a t i
.lutions Manual
5-5
5. Let S r e p r e s e n t a p a s s w o r d t h a t c o n t a i n s S a n d Q .
a n d let W r e p r e s e n t all p o s s i b l e p a s s w o r d s ,
a) T h e n u m b e r of possible p a s s w o r d s containing S
T h e p r o b a b i l i t y t h a t a p a s s w o r d c h o s e n at r a n d o m will
i n c l u d e S a n d Q is ^
, or about 0.002 96 or 0 . 2 9 6 % .
338
a n d Q is 2P2 noPa. o r 2 ! -10^3. O r d e r in p a s s w o r d s is
important, so permutations are used.
6. L e t F r e p r e s e n t t h e f o u r f n e n d s b e i n g o n t h e t e a m ,
fi(S) = 2!
a n d let T r e p r e s e n t all t h e p o s s i b l e t e a m s .
a ) T h e n u m b e r o f w a y s 4 p e o p l e c a n b e p l a c e d in
4 s p o t s v v t ' f n o r d e r is i m p o r t a n t , is 4 P 4 .
n{S)
3
10!
2'
(10-3)!
10!
Jl
n{S) = 2\
n(
10 9 8 7!
n ( S ) = (2-1)
7!
nil :•
n(S) = 2 - 1 0 - 9 - 8
nil
n ( S ) - 1440
-
1
n(i
T h e total n u m b e r o f p a s s w o r d s o f t h i s f o r m is
n{W)
n{W)
n{W)
lePrwPz-
1 • /
n(f I - - I
(262 6 ! 2 ) ! (1010!
- 3)!
T h e n u m b e r of w a y s 9 p e o p l e c a n b e p l a c e d in
4 s p o t s , w h e n o r d e r is i m p o r t a n t , is 9P4,
26!
10!
n(T)=,P,
24!'
7!
2 6 2 5 2 4 ! 1 0 - 9 8 7!
24!
niT)
=
7!
4)1
n-
n(IV) = 2 6 - 2 5 - 1 0 - 9 - 8
n{W)
lO
5!
468 000
n(l)
9-8y7-6-5!
n{T)
9-8 7-6
3024
Now determine the probability.
n{S)
P{S)
n{W)
T h o probability can now be determined.
1440
P{S)
P
468000
n(T)
1
24
P{S)
325
T h e p r o b a b i l i t y t h a t a p a s s w o r d c h o s e n at r a n d o m will
i n c l u d e S a n d Q is
1
3024
J_
126
or a b o u t 0 , 0 0 3 0 8 o r
325
0.308%.
b) S i n c e r e p e t i t i o n is a l l o w e d , t h e n u m b e r of
p a s s w o r d s c o n t a i n i n g S a n d Q is 2P2 - 1 0 ^ or 2 ! - l O l
T h e total n u m b e r o f p a s s w o r d s a v a i l a b l e is 2 6 ^ l o J
P(S)
P(F):
T h e p r o b a b i l i t y t h e 4 f r i e n d s will all b e o n t h e t e a m is
1 , or about 0.0079 or 0 . 7 9 % .
126
b) If o n l y 8 s t u d e n t s w e r e a p p l y i n g f o r t h e t e a m , t h e n t h e
t o t a l n u m b e r of t e a m s w o u l d c h a n g e f r o m 9P4 t o 8P4.
n(r) =
n{W)
sP.
8!
P(S) =
2! • 1 0 '
26^
2
P{S)
676
(8-4)!
8!
niP)
4!
26^
2 1
PiS)
=
2 6 - 10^
2!
P(S)
niT)
8-7-6-5-4!
niT)
=
4!
fi(r) = 8 - 7 - 6 - 5
n(r)
1680
1
P(S)
5-6
338
C h a p t e r 5: P r o b a b i l i t y
N o w determine the probability,
P(F):
n(T)
24
P(F):
1680
1
8. a) L e t Y r e p r e s e n t Y u k o . Luigi a n d J u s t i n b e i n g
c h o s e n , a n d let T r e p r e s e n t all of t h e w a y s t h a t
t r e a s u r e r , s e c r e t a r y a n d l i a i s o n c a n b e c h o s e n . In t h i s
c a s e , t h e r e is o n l y 1 w a y to a c h i e v e t h e f a v o u r a b l e
o u t c o m e . T h e n u m b e r o f w a y s to c h o o s e t r e a s u r e r ,
s e c r e t a r y a n d liaison is 15P3.
n{T)
T h e probability increases from - ^ L to
or about
126
70
0 . 0 1 4 3 or 1,43%.
7. Let T r e p r e s e n t T a r a a n d L a u r a b e i n g c h o s e n to p l a y
in t h e infield. Let O r e p r e s e n t all p o s s i b l e infield l i n e u p s .
T h e n u m b e r o f w a y s to a r r a n g e T a r a a n d L a u r a in t h e
infield p o s i t i o n s is 2 • 3L T h e n u m b e r o f w a y s to a r r a n g e
t h e o t h e r 7 p l a y e r s in t h e r e m a i n i n g 2 infield p o s i t i o n s is
'P?- T h e r e f o r e , t h e total n u m b e r o f infield l i n e u p s t h a t
i n c l u d e T a r a a n d L a u r a is (2 • 3!) • 7P2.
n(r) = (2-3!)-,P^
15!
n{T)
(15-3)!
niT) =
15!
12!
1 5 - 1 4 13 121
npr)=
^ '
12'
n(T)
15-14-13
11(7) = 2 7 3 0
N o w determine the probability.
p p f ) = ' m
11(1)
^
'
1
(7-2)!
p{y)=
2730
n(r)-(2-3!)-|
1
T h e p r o b a b i l i t y is
2730
1
n(r) = (2-3-2-1)
5!
n(r) = 2 - 3 - 2 - 7 - 6
n(r) = 504
T h e total n u m b e r o f infield l i n e u p s p o s s i b l e is 9P4,
n(0)
=
( 9 -9!4 ) '
0.0366%.
b) Let Y r e p r e s e n t Y u k o . L u i g i a n d J u s t i n b e i n g
c h o s e n , a n d let L r e p r e s e n t all o f t h a t w a y s t h a t t h r e e
s t u d e n t s c a n b e c h o s e n . A g a i n , t h e r e is o n l y 1 w a y to
a c h i e v e t h e f a v o u r a b l e o u t c o m e . T h e n u m b e r of w a y s
to c h o o s e 3 p e o p l e f r o m 15 t o c l e a n u p is 15C3.
C,
n(L) =
15!
niL) =
( 1 5 - 3 ) ! • 3!
9!
15!
5!
n{0)
12!
9-8-7-6-5!
n(0)
5!
n(L)-~
12?
3024
n(T)
P(T)
niP)
=
504
3024
1
it^H
nil}
N o w determine the probability,
P(T)
15 U
^9-8-7-6
r?(0)
or a b o u t 0 . 0 0 0 3 6 6 or
n{L) =
n(L)
13 12!
;
' 1
13
32""""
2730
455
N o w determine the probability.
PiY)=Bm
n{L)
T h e p r o b a b i l i t y t h a t T a r a a n d L a u r a will b o t h p l a y in
t h e infield is | . T h e r e f o r e t h e o d d s in f a v o u r o f t h i s
o
e v e n t a r e 1 : ( 6 1 ) or 1 : 5,
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
T h e p r o b a b i l i t y t h a t Y u k o . L u i g i a n d J u s t i n will be
p i c k e d IS
455
. or a b o u t 0 0 0 2 2 or 0 . 2 2 0 % .
5-7
9. a ) Let F r e p r e s e n t a p a s s w o r d t h a t is g r e a t e r t h a n
5 0 0 0 . a n d let O r e p r e s e n t all p o s s i b l e p a s s w o r d s .
T h e r e a r e 4 9 9 9 4^digit n u m b e r s g r e a t e r t h a n 5 0 0 0 .
H o w e v e r , in t h i s r a n g e , t h e r e a r e 5 n u m b e r s t h a t
r e p e a t all 4 d i g i t s ( 5 5 5 5 . 6 6 6 6 , 7 7 7 7 , 8 8 8 8 a n d 9 9 9 9 ) .
Therefore, the total n u m b e r of usable p a s s w o r d s ,
g r e a t e r t h a n 5 0 0 0 , is 4 9 9 4 . T h e t o t a l n u m b e r of f o u r
d i g i t s p o s s i b l e is 10 0 0 0 , H o w e v e r , t h e r e a r e 10
n u m b e r s t h a t r e p e a t all 4 d i g i t s . S o . t h e r e a r e 9 9 9 0
usable passwords,
T h e probability that Lesley's password begins with an
250
o d d digit a n d e n d s w i t h a n e v e n digit is ^ , o r
999
0.2503 or 2 5 , 0 3 % ,
10. L e t T r e p r e s e n t t h r e e giris a n d t w o b o y s b e i n g
c h o s e n t o f o r m a s u b c o m m i t t e e , a n d let S r e p r e s e n t
all p o s s i b l e s u b c o m m i t t e e s
In t h i s e x a m p l e , o r d e r is n o t i m p o r t a n t . T h e n u m b e r of
w a y s t o a r r a n g e t h r e e giris a n d t w o b o y s f r o m 16 girls
a n d 7 b o y s is nCz - 7C2.
mm
16!
11(7) =
P{F)
4995
T h e p r o b a b i l i t y L e s l e y ' s p a s s w o r d is g r e a t e r t h a n
5 0 0 0 is
2497
or a b o u t 0 , 4 9 9 9 o r 4 9 . 9 9 % .
4995
b) L e t B r e p r e s e n t a p a s s w o r d t h a t b e g i n s a n d e n d s
w i t h t h e n u m b e r 4 , a n d let O r e p r e s e n t all p o s s i b l e
p a s s w o r d s . B e c a u s e it is a s s u m e d t h a t t h e first a n d last
digits are 4s, only the t w o middle digits n e e d to be dealt
w i t h . S i n c e r e p e a t i n g t w o d i g i t s is a l l o w e d , t h e n u m b e r
of w a y s t o a r r a n g e t h e t w o m i d d l e d i g i t s is W\ o r 1 0 0 ,
H o w e v e r , all t h e d i g i t s c o u l d b e 4 . w h i c h is not a l l o w e d .
Therefore, the total n u m b e r of usable p a s s w o r d s ,
w h e r e t h e first a n d last d i g i t s a r e 4 s , is 9 9 . A g a i n , t h e
t o t a l n u m b e r of u s a b l e p a s s w o r d s is 9 9 9 0 .
about
n{T)
7!
( 1 6 - 3 ) 1 • 3! ( 7 - 2 ) ! 2!
_ 16! _
_ 7 L
13! • 3 ! ' 5 ! ' - 2 !
16-15-14-13!
n{T) =
13! • 3 2 1
7 6 5!
2 1
5!
1 6 - 1 5 14 7 6
n(T) =
n(7)
5 6 0 21
n ( 7 ) = 11 7 6 0
T h e n u m b e r o f w a y s t o a r r a n g e 2 3 p e o p l e in a f i v e p e r s o n c o m m i t t e e is 23C5.
n{S)
niS) =
23!
( 2 3 - 5 ) ! • 5!
23!
n{S)
1 8 ! • 5!
23-22-2T20-19-18!
P{B)
n{S)
1110
w i t h 4 is
1I
18!
n{S)
5 4
o r a b o u t 0 , 0 0 9 9 or 0 , 9 9 % .
110
c ) Let E r e p r e s e n t a p a s s w o r d t h a t b e g i n s w i t h a n o d d
n u m b e r a n d e n d s w i t h a n e v e n n u m b e r , a n d let O
r e p r e s e n t all p o s s i b l e p a s s w o r d s . If t h e first digit is
o d d a n d t h e last digit is e v e n , t h e n t h e r e is n o
p o s s i b l e w a y t h a t all f o u r d i g i t s c a n b e t h e s a m e . T h e
first digit c a n b e o n e of f i v e o d d n u m b e r s ( 1 , 3, 5, 7
a n d 9 ) , a n d t h e last digit c a n b e o n e o f 5 e v e n
n u m b e r s ( 0 , 2, 4 , 6 a n d 8 ) . T h e t w o m i d d l e d i g i t s c a n
b e o n e of t e n d i g i t s . T h e r e f o r e t h e t o t a l n u m b e r of
u s a b l e p a s s w o r d s t h a t b e g i n w i t h a n o d d digit a n d
e n d w i t h a n e v e n digit is 5 10 10 5. o r 2 5 0 0 . T h e
total n u m b e r of u s a b l e p a s s w o r d s is 9 9 9 0 ,
n{E)
P{E)
n{0)
5 13
2
r
23 2 2 21 20 19
T h e probability I esley's password begins and e n d s
3 2
4 037 880
n{S) =
n{S)
120
33 649
N o w determine the probability.
P{T)
mil
n(S)
11760
PiT)
33649
1680
P{T)
4807
T h e o d d s in f a v o u r t h a t t h e c o m m i t t e e will c o n t a i n
3 giris a n d 2 b o y s is 1 6 8 0 : 4 8 0 7 - 1 6 8 0 o r
1680 : 3127.
2500
P{E)
9990
250
P{E)
5-8
999
C h a p t e r 5: P r o b a b i l i t y
5 is 2^, o r 16, T h e r e
5 l a n d o n tails (all
.bability t h a t at l e a s t
N o w determine the probability.
m
PiL)
n(S)
u !i375 or 9 3 . 7 5 % ,
1 2 . Let e r e p r e s e n t B i l y a n a a n d B o j a n a sitting
t o g e t h e r , a n d let S r e p r e s e n t all p o s s i b l e s e a t i n g
arrangements.
a) T h e n u m b e r o f w a y s to s e a t B i l y a n a a n d B o j a n a
t o g e t h e r is 4 • 2P2, •
T h e n u m b e r of w a y s t o s e a t
t h e 3 o t h e r p e o p l e is 3P3, o r 3!, T h e n u m b e r of w a y s t o
s e a t t h e 5 f n e n d s in t h e r o w is 4 • 2 ! • 3!, o r 4 8 .
T h e r e a r e 5! or 1 2 0 w a y s t o a r r a n g e five p e o p l e .
P{B)
n{S)
Pisy:
^«
120
2
P{B)
P ( L ) = 120
1680
1
P(l) = ^^
14
T h e p r o b a b i l i t y that t h r e e of t h e f o u r c o u r s e s will b e
p s y c h o l o g y , linear a l g e b r a , a n d E n g l i s h is ^
a b o u t 0 , 0 7 1 4 or 7 , 1 4 % ,
b) Let R r e p r e s e n t T a n y a t a k i n g r e l i g i o n , political
s t u d i e s a n d b i o l o g y in h e r first t e r m , a n d let S
r e p r e s e n t all p o s s i b l e s c h e d u l e s .
T h e r e are 10 3!, or 6 0 , w a y s to a r r a n g e r e l i g i o n ,
political s t u d i e s a n d b i o l o g y . T h e n u r n b e r of w a y s t o
fill t h e r e m a i n i n g t w o slots is 5P2n ( P ) = 60 c P.
5
T h e p r o b a b i l i t y t h a t B i l y a n a a n d B o j a n a a r e sitting
t o g e t h e r is ^ , 0,4 or 4 0 % .
5
b) T h e p r o b a b i l i t y t h a t B i l y a n a a n d B o j a n a a r e n o t
sitting t o g e t h e r is t h e c o m p l e m e n t of t h e p r o b a b i i i t y
t h a t t h e y a r e sitting t o g e t h e r
P(BO= 1
P{B)
P{B)
= 1-
P{B)
=
I
\
o
T h e p r o b a b i l i t y B i l y a n a a n d B o j a n a a r e not sitting
t o g e t h e r is - , 0.6 o r 6 0 % .
5
13. a) Let L r e p r e s e n t T a n y a t a k i n g p s y c h o l o g y , linear
a l g e b r a a n d E n g l i s h in h e r first t e r m , a n d let S
r e p r e s e n t all p o s s i b l e s c h e d u l e s .
T h e r e a r e 4 • 3!, o r 2 4 w a y s t o a r r a n g e p s y c h o l o g y ,
linear a l g e b r a a n d E n g l i s h . T h e n u m b e r o f w a y s to fill
t h e f o u r t h slot is 5. T h e r e f o r e , t h e n u m b e r of
schedules that contain psychology, linear algebra and
E n g l i s h is 2 4 5, or 1 2 0 .
niR)
=6
niR)
=60.^
3!
niR)
= 6(-
T h e total n u m b e r o f w a y s t o o r g a n i z e 8 c o u r s e s into
5 s l o t s IS 8P5.
niS) = ..P
8!
n{S)
(8-5)!
8!
3!
8 - 7 - 6 - 5 - 4 - 3!
niS)
3!
n(S) = 8 - 7 - 6 - 5 - 4
n(S) - 6720
N o w determine the probability.
n{R)
PiR)
niS)
1200
4)!
=
6720
5
PiR)
8!
28
T h e p r o b a b i l i t y t h a t t h r e e of t h e five c o u r s e s will b e
4!
niS) =
'
n{R) = 1200
PiR)
8!
(8
. ^
(5-2)!
3!
.P,
n{S)
0
60-5 4
n{R)
T h e total n u m b e r o f w a y s t o o r g a n i z e 8 c o u r s e s into
4 s l o t s is 8P4.
n(S)
, or
8 - 7 - 6 5 4!
r e l i g i o n , political s t u d i e s a n d b i o l o g y is
4!
n(S):
8 7 6 5
niS)
1680
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
. or about
28
0.179 or 17.9%.
5-9
14. a) Let A r e p r e s e n t a h a n d t h a t c o n t a i n s t h e a c e to 8
of t h e s a m e suit, a n d let O r e p r e s e n t all 8 - c a r d h a n d s .
In this e x a m p l e , o r d e r is not i m p o r t a n t . T h e r e a r e
4 w a y s to h a v e t h e a c e to 8 o f t h e s a m e suit, b e c a u s e
t h e r e a n - I s u i t s . T h e r e a r e ssCa w a y s to a r r a n g e
5 2 c a r d s in 8 p o s i t i o n s .
N e w df4...'j.irh- lha: <-,rr,n .bility.
P(f>)
Off))
4 .4 19
7 0 I/
P('--)
=
P{-S')
44! • 8!
4 ¥ J ¥ T ¥ ^ 5 ^ 4 d r 2 l
•< o
f
••
4
t
PiA]
n(0)
22. 4 / .SO
y 2
/
4/
209
P(S) = ^ '
50 337
T h e p r o b a b i l i t y t h a t all c a r d s in a n 8 - c a r d h a n d a r e t h e
c ) Let F r e p r e s e n t a h a n d c o n t a i n i n g 4 f a c e c a r d s a n d
4 o t h e r c a r d s , a n d let O r e p r e s e n t all 8 - c a r d h a n d s ,
A s t a n d a r d d e c k o f c a r d s c o n t a i n s 12 f a c e c a r d s a n d
4 0 o t h e r c a r d s . T h e r e a r e 12C4 w a y s to c h o o s e 4 f a c e
c a r d s , a n d 40C4 w a y s t o c h o o s e 4 o t h e r c a r d s .
T h e r e f o r e , t h e n u m b e r o f w a y s t o h a v e 4 face c a r d s
a n d 4 o t h e r c a r d s in a h a n d is 1 2 C 4 • 4 0 C 4 .
niA)
/• •} I ] if
11
t/
s a m e c o l o u r is —
, or a b o u t 0 . 0 0 4 1 5 or 0 , 4 1 5 % ,
50337
N o w determine the probability.
PiA;
50
2
n(O)-7-9-13-17-23-47-50
P{A)
19
7
52-5T 50-49-48•47-46-45-44!
,(0).ii.l9.4§.52,51.46,^^,5Q
^
1/
/ 9 17 4 / 2 5 5
52!
n{0)
bt)
V
4 5 U) :> 11
(52-8)1-8!
n(0) =
14 22 2:i
1.3
/ 0 13 1 7 " 2 4 /
Pf.4)
'
') .')
pp:.)
52'
'
n(S,
^
! ? 24 4 / ^ 2
/o
2
PiA)
12!
/ 0 I 1 1 / 22 4./ 2 5
P{A)
n{P)
2 g 9 'o75
n(f)
T h e p r o b a b i l i t y t h a t a n e i g h t - c a r d h a n d will c o n t a i n t h e
A , 2. 3, 4 . 5, 6, 7 a n d 8 of the s a m e suit is
?
376269075
nil)
n(P)
n(S) = 2 - , , C ,
4)1 • 4 ! ( 4 0 - 4 ) ! • 4 !
12'
H'
401
4! 36!
4!
12 11 10 0 8" 4 0 4 9 38 37 3 6 !
or a b o u t 0 . 0 0 0 0 0 0 0 0 5 3 2 .
b) Let S r e p r e s e n t a h a n d c o n t a i n i n g 8 c a r d s of t h e
s a m e c o l o u r , a n d let O r e p r e s e n t all 8 - c a r d h a n d s .
T h e n u m b e r of w a y s to arrange 8 cards f r o m 26 cards
of t h e s a m e c o l o u r suit is aeCs. But. b o t h c o l o u r s a r e
c o n s i d e r e d . S o , t h e n u m b e r of w a y s t o h a v e 8 c a r d s
o f t h e s a m e c o l o u r is 2 - zeCa.
(12
40!
n(P)
n(F)
81
4 3 2 1
36! • 4 - 3 - 2 - 1
12 11 10 9 4 0 3 9 3 8 37
•1 4. 2
f \?^
10
' 4 3 2
"""4^-3-2""""
9 11
40,55 45.37
4 3
2
5 0 10 11 13 19 3 7
N o w determine the probability,
P(F)
n(P)
niO)
.(S).2-
n(S)
( 2 6 - 8 ) ! • 8!
Pit)
26!
2
P(P)
2-26-25-24-23 •22-21-20-19-18!
^^TTiT7:6":5:4^3:2l
n{S)
n{S)
n{S)
5-10
P(P)
26-25-24-23-22-21-20-19
8J776T5T4T3
n{S)
3^
5 10 11-19 37
18! • 8!
n{S)
5 - 9 10 11 13 19
7 9 13 17 2 3 4 7 5 0
24
21
25
20
tt 3
7 3
5
4
5 5 13 10 2 2 2 3
26
" 1 9
2
22 2 3
11 19 37
7 T 7 23 47
7733
PiF)
26-25-24-23-22-21-20-19
8-7-3-2-5-4-3
7 77^-23 4 7 5 10
128 6 3 9
T h e probability that an 8-card hand contains 4 face
c a r d s a n d 4 o t h e r c a r d s is y ^ ^ | ^ g • or a b o u t 0 , 0 6 0 1
or 6 , 0 1 % ,
C h a p t e r 5: P r o b a b i l i t y
1 5 . Let A r e p r e s e n t a 3 4 e t t e r w o r d , u s i n g t h e l e t t e r s in
the word "CABINET", that contains two vowels and one
c o n s o n a n t . Let O r e p r e s e n t all 3 d e t t e r w o r d s t h a t u s e t h e
letters in t h e w o r d " C A B I N E T " . T h e w o r d " C A B I N E T "
contains 3 vowels and 4 consonants. Therefore, there
a r e 3C2 • 4C1 w a y s t o a r r a n g e 2 v o w e l s a n d 1 c o n s o n a n t
from 3 vowels and 4 consonants.
3!
niA) =
(3^2)!3!
niA)
niA)
I
)! -1!
4!
1! -2! 3 ! - 1 !
i • < !-•• ;• •; -i'lty t h a t t h e b e a r d e d collie a n d t h e s h e l t i e
3-2! 4-3!
will b e n e x t t o e a c h o t h e r is | , 0 2 5 or 2 5 % ,
2!
3!
n(4) = 3-4
1 7 . Let A r e p r e s e n t M a g g i e a n d T a n y a in t h e s t a r t i n g
lineup.
n ( 4 ) = 12
T h e n u m b e r o f p o s s i b l e a r r a n g e m e n t s of 3 l e t t e r s
f r o m 7 letters is 7C3.
4!
! - 2! ( 4 - 1 ) 1 - 1 !
7!
11(0) =
(7-3)!
niO) =
3!
3 ! -1!
7!
•
•••1!
423!
4 ! -3!
n(0) =
3!
7-6-5-4!
41 2 „^
niO) =
7 6 5
i-l A:
-
iid
3 2
niO} =
210
b<-
)f w a y s in w h i c h M a g g i e a n d T a n y a c a n
l i n e u p is 6 0 ,
:h.
' s . r i i n g
6
2„
n ( 0 ) = 35
b' N o w determine the probability.
PiA)
'
3,! . 3! < 1
n(0) =
=
-
2!
5!
niO)
12
PiA)
17(0) =
35
7" 3 2
2
T h e p r o b a b i l i t y t h e tiles will c o n s i s t o f 2 v o w e l s a n d
1 c o n s o n a n t is
12
35
3 z
, or a b o u t 0,343 or 3 4 . 3 % ,
n{0)
z
350
T h e t o t a l n u m b e r of p o s s i b l e s t a r t i n g l i n e u p s is 3 5 0 .
16. Let D r e p r e s e n t a n a r r a n g e m e n t of t h e e i g h t d o g s
s u c h t h a t t h e b e a r d e d collie a n d t h e s h e l t i e a r e
t o g e t h e r . Let O r e p r e s e n t all a r r a n g e m e n t s .
T h e n u m b e r o f w a y s t o p l a c e t h e b e a r d e d collie a n d
t h e s h e l t i e t o g e t h e r is 7 • 2!. T h e n u m b e r of w a y s t o
p l a c e t h e o t h e r 6 d o g s is 6!. T h e r e f o r e , t h e total
n u m b e r of w a y s t o s e a t all t h e d o g s , w i t h t h e collie
a n d s h e l t i e t o g e t h e r , is 7 2 ! - 61, o r 2 • 71.
T h e r e a r e 8! w a y s t o a r r a n g e e i g h t d o g s .
niA)
PiA)
PiA) =
PiA)
n{0)
^
350
35
T h e p r o b a b i l i t y b o t h M a g g i e a n d T a n y a will b e in t h e
s t a r t i n g l i n e u p is - ~ , or a b o u t 0 , 1 7 1 o r 1 7 . 1 % .
35
F o u n d a t i o n s of M a t h e m a t i
liutions Manual
5-11
18. e . g . , I w o u l d u s e p e r m u t a t i o n s in a p r o b l e m w h e r e
the order of the items w a s important, and use
c o m b i n a t i o n s in a p r o b l e m w h e r e o r d e r w a s n o t
important. Permutations; Determine the probability
t h a t t w o i t e m s a r e n e x t t o e a c h o t h e r in a l i n e u p of
s e v e n d i f f e r e n t i t e m s t h a t h a s b e e n p l a c e d in a
random order. Combinations: Determine the
probability that, given eight books, four of w h i c h are
a b o u t m a t h , if I c h o o s e f i v e of t h e e i g h t b o o k s . I
choose three math books.
1 - ^aiiim
365*
0 . 7 0 6 . . . . T h i s a n s w e r w a s a b o u t 7 1 % , so I
t h o u g h t n w o u l d b e a little m o r e . I t r i e d n = 3 5
1 ^ J i 5 5 i | = 0 . 8 1 4 . . . T h i s is a little o v e r 8 0 % . T h e r e ^
365-^'
w o u l d n e e d t o b e 3 5 p e o p l e in a r o o m f o r t h e
p r o b a b i l i t y t h a t t w o o f t h e m h a v e t h e s a m e b i r t h d a y is
80%
History Connection, page
1 9 . Let A r e p r e s e n t a r o u t e t h a t p a s s e s t h e p o o l , a n d
let O r e p r e s e n t all r o u t e s . T h e r o u t e T u y e t t a k e s
c o n s i s t s of 5 ' r i g h t s ' a n d 4 d o w n s ' . T h e r e f o r e , t h e
t o t a l n u m b e r o f p o s s i b l e r o u t e s is t h e t o t a l n u m b e r of
m o v e s (9!), dividing out repetition (5! • 4!)
91
n(0) =
5'
4'
9
n(0) =
n{0)
11(0):
n{0)
5'
1
9
4 2
31)/
24
126
T o p a s s by t h e p o o l , T u y e t m u s t first t a k e 3 ' r i g h t s '
and 2 'downs', then must take 2 'nghts' and 2 downs'
in a n y o r d e r . A g a i n , r e p e t i t i o n m u s t b e d i v i d e d o u t .
5!
n{A)
3!
4!
2!
2! 2!
5 - 4 3!
n{A)
4
3! • 2 1 2!
5-4
2
3-2!
2 1
4-3
•
2
n{A)
5 2 2 3
n{A)
60
N o w determine the probability.
n{A)
P{A)
-
n(0)
60
324
L e t u s n o w f o l l o w in P a s c a l ' s f o o t s t e p s a n d a n a l y z e
c o r r e c t l y t h e c h a n c e s o f w i n n i n g in t h e s e t w o g a m e s .
S i n g l e d i e R o l l i n g a s i n g l e die o n c e l e a d s t o
p r e c i s e l y o n e o f 6 p o s s i b l e o u t c o m e s : E x a c t l y o n e of
t h e n u m b e r s 1 , 2, 3, 4 . 5, o r 6 will b e o n t o p . T h e die
is c a l l e d fair, if e a c h of t h e s e o u t c o m e s is e q u a l l y
likely. P l a y e r s of d i c e g a m e s u s u a l l y a s s u m e t h e y a r e
u s i n g fair d i c e , s o I will a s s u m e t h i s t o o . If I roll a die
4 t i m e s , t h e n t h e t o t a l n u m b e r of all p o s s i b l e
o u t c o m e s is 6 6 6 6 - 1 2 9 6
Out of these there are 5 5 5 5 = 6 2 5 o u t c o m e s
w i t h n o 6 in t h e m . T h u s , if I b e t o n g e t t i n g a t l e a s t o n e
6 w h e n rolling a d i e 4 t i m e s , t h e r e a r e 6 2 5
p o s s i b i l i t i e s of l o s i n g , a n d 1 2 9 6 - 6 2 5 = 6 7 1
p o s s i b i l i t i e s of w i n n i n g . S o . m y c h a n c e s of w i n n i n g
a r e h i g h e r t h a n m y c h a n c e s of l o s i n g .
T w o d i c e Let us n o w turn to the two-dice g a m e .
R o l l i n g t w o d i c e o n c e l e a d s t o o n e of 3 6 p o s s i b l e
o u t c o m e s , n a m e l y all p o s s i b l e o u t c o m e s of rolling d i e
n u m b e r 1 c o m b i n e d w i t h all p o s s i b l e o u t c o m e s of
rolling d i e n u m b e r t w o . T h u s , if w e roll t w o d i c e
2 4 t i m e s , t h e n t h e total n u m b e r o f p o s s i b l e o u t c o m e s
is 36 3 6 ... 3 6 ( 3 6 m u l t i p l i e d w i t h itself 2 4 t i m e s ) ,
w h i c h is a p p r o x i m a t e l y 2 , 2 4 5 . . . • 1 0 ^ ' . O u t of t h e s e
t h e r e a r e 3 5 35
,, 3 5 ( 3 5 m u l t i p l i e d w i t h itself
2 4 t i m e s ) , w h i c h is a p p r o x i m a t e l y 1 . 1 4 1 , ., • 10^^
o u t c o m e s w i t h n o d o u b l e 6. T h u s , if I g a m b l e o n
g e t t i n g at l e a s t o n e d o u b l e 6 w h e n rolling t w o d i c e
24 times, there are approximately 1,141,., • 10'''
p o s s i b i l i t i e s of l o s i n g , a n d 1 , 1 0 3 . , , • 10^^ p o s s i b i l i t i e s
of w i n n i n g . This m e a n s that the c h a n c e s of winning
with this g a m e are lower than the c h a n c e s of l o s i n g —
as the Chevalier De Mere learned the hard way,
10
P{A)
21
10
T h e p r o b a b i l i t y t h a t T u y e t p a s s e s t h e p o o l is - - . or
Mid-Chapter Review, page 327
1. a) O u t c o m e t a b l e
D i c e roll
Product
Sum
Who wins?
(1=1.1]
(1,1,2)
(1,1,3)
1
3
Enk
2
4
I d Ik
3
5
Enk
_I1,1,4)
JT2JJ
4
6
brik
2
4
Erik
n e e d t o b e a lot o f p e o p l e f o r t h e p e r c e n t a g e t o b e a s
(1,2,2)
4
5
fnk
h i g h a s 8 0 % . I tried n = 1 0 0 .
(1,2,3)
6
" 6
J4,2,4)
8
7
Ethan
(1,3,1)
(1,3,2)
3
5
Enk
(1,3,3)
_ 9
6
r ,
47.6%.
2 0 . 1 k n o w f r o m c h a p t e r 4 t h a t t h e f o r m u l a t o u s e is
1
, w h e r e n is t h e n u m b e r of p e o p l e . I t h e d a
365"
large number, b e c a u s e I thought that there would
P
1 - 365 100 j h e c a l c u l a t o r g a v e a n e r r o r m e s s a g e .
365'*
T h i s n u m b e r w a s t o o h i g h . I tried n = 3 0 .
5-12
^
1
Ethan
Ethan
Ethan
C h a p t e r 5: P r o b a b i l i t y
(1,3,4)
12
8
Ethan
(1,4,1)
(1.4,2)
4
6
Enk
8
7
Ethan
(1,4,3)
12
8
Ethan
(1,4,4)
16
9
Ethan
(2,1,1)
(2,1,2)
2
4
Enk
4
5
Enk
(2,1,3)
6
6
Ethan
(2,1,4)
8
7
Ethan
2
3
4
5
(2,2,1)
4
5
Enk
3
4
5
6
7
(2,2,2)
8
6
Ethan
4
5
6
7
8
(2,2,3)
12
7
Ethan
5
6
7
8
9
(2,2,4)
16
8
Ethan
(2,3,1)
6
6
Ethan
(2,3,2)
12
7
Ethan
(2,3,3)
18
8
Ethan
(2,3,4)
24
9
Ethan
(2,4,1)
8
7
Ethan
(2,4,2)
16
8
Ethan
(2,4,3)
24
9
Ethan
(2,4,4)
32
10
Ethan
(3,1,1)
(3,1,2)
3
5
Enk
6
6
Ethan
(3,1,3)
9
7
Ethan
(3,1,4)
12
8
Ethan
(3,2,1)
6
Ethan
(3,2,2)
12
6
7
(3,2,3)
18
8
Ethan
(3,2,4)
24
9
Ethan
(3,3,1)
9
7
Ethan
(3,3,2)
18
8
Ethan
(3,3,3)
27
9
Ethan
(3,3,4)
36
10
Ethan
Ethan
(3,4,1)
12
8
Ethan
(3,4,2)
24
9
Ethan
(3,4,3)
36
10
Ethan
(3,4,4)
48
11
Ethan
(4,1,1)
(4,1,2)
4
6
Enk
8
7
Ethan
(4,1,3)
12
8
Ethan
(4,1,4)
16
9
Ethan
(4,2,1)
8
7
Ethan
(4,2,2)
16
8
Ethan
(4,2,3)
24
9
Ethan
(4,2,4)
32
10
Ethan
(4,3,1)
12
8
Ethan
(4,3,2)
24
9
Ethan
(4,3,3)
36
10
Ethan
(4,3,4)
48
11
Ethan
(4,4,1)
16
9
Ethan
(4,4,2)
32
10
Ethan
(4,4,3)
48
11
Ethan
(4,4,4)
64
12
Ethan
P ( s u m is l i i g l i e r ) =
b) F a i r e . g . , C o i n s h a v e a n e q u a l c h a n c e o f l a n d i n g
o n heads and landing o n tails. Therefore, since e a c h
player has the s a m e requirements to w i n , but with the
o p p o s i t e c o i n f a c e , t h i s g a m e is fair.
2. a) O u t c o m e T a b l e ;
S inner 1
Cl
CO
6
T o t a l n u m b e r of o u t c o m e s = 2 0
b) i) P ( s u m o f 5 ) =
20
P ( s u m of 5) =
^_
5
T h e p r o b a b i l i t y o f g e t t i n g a s u m of 5 is - , 0.2 o r 2 0 % .
5
3
ii) P ( s u m of 8 o r 9 ) = — T h e p r o b a b i l i t y o f g e t t i n g a
3
s u m of 8 or 9 is — , 0 . 1 5 o r 1 5 % .
20
3
iii) P ( s u m o f 4 ) = —
T h e p r o b a b i l i t y of g e t t i n g a s u m
of 4 is — , 0 . 1 5 o r 1 5 % .
20
3. a) T h e r e a r e 2 6 b l a c k c a r d s in a s t a n d a r d 5 2 - c a r d d e c k .
P(black card)
26
52
P(black card)
2
T h e p r o b a b i l i t y of d r a w i n g a b l a c k c a r d is ^ , 0.5 o r 5 0 % .
b) S i n c e 2 6 c a r d s a r e b l a c k , t h e n 2 6 c a r d s a r e r e d .
T h e r e f o r e , t h e o d d s in f a v o u r o f t h e c a r d b e i n g b l a c k
are 2 6 ; 26, or 1 ; 1.
c ) T h e r e a r e 13 d i a m o n d s in a s t a n d a r d d e c k , s o
there are 39 cards that are not d i a m o n d s . Therefore,
the odds against the card being a diamond are
3 9 ; 13, o r 3 : 1 .
d) F o r t y c a r d s in e a c h d e c k a r e not f a c e c a r d s .
P ( n o t face card):
5'^^
2
10
P ( n o t face card) = -
T h e p r o b a b i l i t y t h e c a r d d r a w n is not a f a c e c a r d is
10
13 '
or about 0.769 or 7 6 . 9 % .
13
64
P ( s u m is not h i g l i e r ) =
52
64
T h i s g a m e is not fair. E t h a n h a s t h e a d v a n t a g e . T h e r e
are more products that are greater than the s u m s .
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5-13
/
4.
Plwin:
;i) 1 .'t /• ..-pj.'K.M.i n h.liid {.(jtiC^iioruj ihc
Umn,
qutM-n ,ind |a(-k o f die s.nino m i i ! . nnd lot i ) ro()iosf>nl
all f,'U< hn: bonds
Ihorr
art; 4 ddfoie-nt way:> tf) plar.o
I I h ; A , K, <) and ,i of the name; :,uit in a hand (one? lot
Plwini
oaf;h Sllll)
I I m ; h ; ,ire 2 0 wnyo of plat.mg tin- hflh canJ
( ( . h h ; for f'orh rfjrnoining rani)
T h e p r o b a b i l i t y h e will w i n is
. o r a b o u t 0 J 7 9 o r 17 J % .
28
5. a ) T h e p r o b a b i l i t y t h a t t w o o f o u r p r i m e m i n i s t e r s
will h a v e t h e s a m e b i r t h d a y is s l i g h t l y l e s s t h a n 5 0 % ,
ou
thore nro 8 0 ways
to h.ivo' !h>; ,!(..•, kinu oncon. nnd jnok of IheMill in <i hand
o r - • . T h e r e f o r e , t h e o d d s in f a v o u r o f t w o p r i m e
2
ministers having t h e s a m e birthday are slightly less
niO)
than 1 : 1.
b) T h e o d d s a g a i n s t t w o p r i m e m i n i s t e r s h a v i n g t h e
s a m e birthday a r e slightly m o r e t h a n 1 : 1 .
c) Y e s , Sir John A . Macdonald a n d Jean Chretien
both share a birthday o n January 1 1 . However.
M a c d o n a l d ' s official d a t e o f birth r e c o r d e d in t h e
G e n e r a l R e g i s t e r O f f i c e in E d i n b u r g h . S c o t l a n d i s
January 10. But, M a c d o n a l d celebrated his birthday
on January 1 1 .
11(0) =
f hf< total numtK.-i of hands is
i.-vm;
.
24!
(24
5)! - 5!
24!
19! • 5!
n(n
24 2 3 2 2 : 4 2 0 10»
10! 5 4 2
1
2 1 2.t •?? 2 T 2 0
1 2 ?'
100 4 2 ( i
120
n(0) = 42 504
6. L e t A r e p r e s e n t F r o d o , S a m , a n d A r a g o r n b e i n g
chosen for president, treasurer and secretary. Let O
r e p r e s e n t ali p o s s i b l e s e l e c t i o n s . T h e n u m b e r o f w a y s
t h a t F r o d o . S a m . a n d A r a g o r n c a n b e c h o s e n is 3P3 o r
31. T h e n u m b e r o f w a y s t h a t t h e 14 p e o p l e c a n b e
c h o s e n f o r t h e 3 p o s i t i o n s is 14P3.
N o w determine t h e probability.
niA)
PiA)
n(0)
80
PiA)
42504
10
PiA)
5313
T h e p r o b a b i l i t y t h a t a d e a l t h a n d will c o n t a i n a n a c e ,
n{A
n(A)
14!
k i n g , q u e e n a n d j a c k o f t h e s a m e s u i t is
!4 r "
1^' i j !
^ , 1 1
n{A)
14 13 12
n{A) - 2 1 8 4
N o w determine the probability.
P{A)
10
5313
or
about 0.001 8 8 2 or 0.1882%.
b) L e t F r e p r e s e n t a h a n d c o n t a i n i n g f i v e c a r d s o f t h e
s a m e c o l o u r , a n d let O r e p r e s e n t all e u c h r e h a n d s .
T h e r e a r e 1 2 r e d c a r d s a n d 12 b l a c k c a r d s . T h e n u m b e r o f
w a y s a h a n d c a n h a v e e i t h e r c o l o u r is 12C5. C o n s i d e n n g
both colours, t h e n u m b e r of w a y s a dealt hand can have
f i v e c a r d s o f t h e s a m e c o l o u r is 2 • 12C5.
niO)
3!
P{A)
2184
3 2 1
P{A)
2184
6
PiA)
PiA)
2184
1
364
T h e p r o b a b i l i t y F r o d o , S a m , a n d A r a g o r n will b e
c h o s e n is
, or about 0.002 75 or 0.275%.
364
n(P)
T h e total n u m b e r o f h a n d s is 24C5, o r 4 2 5 0 4 .
5-14
C h a p t e r 5: P r o b a b i l i t y
9. L e t C r e p r e s e n t S t e l l a d r o p p i n g 2 l o o n i e s a n d o n e
o t h e r c o i n . L e t O r e p r e s e n t all o f t h e c o m b i n a t i o n s o f
0(0)
3 coins possible.
1584
PiF)
Stella h a s 6 loonies a n d 6 other coins. T h e n u m b e r of
42504
w a y s that 2 loonies a n d 1 other c o m could be dropped
6
tSeCa-eCi,
161
niC)
T h e p r o b a b i l i t y t h a t a d e a l t h a n d c o n t a i n s five c a r d s of
t h e s a m e c o l o u r is
6
6!
niC)
or about 0.037 2 6 7 or 3 , 7 2 7 % .
161
c) Let F represent a hand containing a four of a kind.
6!
n(C) =
4!
a n d let O r e p r e s e n t all e u c h r e h a n d s .
T h e r e a r e 6 d i f f e r e n t w a y s t o h a v e f o u r o f a k i n d in
niC) =
t o p l a c e t h e fifth c a r d T h e r e f o r e , t h e r e a r e 1 2 0 w a y s
ti 5
T h e t o t a l n u m b e r o f h a n d s is 24C5, o r 4 2 5 0 4 .
5 ' 1!
6 5 4!
4'
niC) =
6!
2!
euchre (one for e a c h rank of card). T h e r e a r e 2 0 w a y s
t o h a v e a f o u r o f a k i n d in a d e a l t h a n d .
6!
( 6 - 2 ) ! • 2! ( 6 - 1 ) ! - 1 !
6 5!
2 1
s ! ^
/i(C) = 3 - 5 - 6
niC) = 90
n(0)
T h e total n u m b e r of w a y s that 3 coins c a n b e d r o p p e d
PiF)
=
120
;
42504
IS 12C3.
n(0) =
12!
1771
niO)
= ( 1 2 - 3 ) ! -3!
T h e p r o b a b i l i t y t h a t a d e a l t h a n d will h a v e f o u r o f a
5
k i n d is
or about 0.002 8 2 3 or 0 , 2 8 2 % .
12!
7y P
n(0)^
1771
8. L e t A r e p r e s e n t a playlist in w h i c h E m a n u e l l a ' s six
n
n{0)
9! •3-2-1
favourite s o n g s are played together. Let O represent
T h e number of w a y s to arrange Emanuella's favourite
n(P)
T h e n u m b e r o f w a y s t o a r r a n g e t h e o t h e r 2 4 s o n g s is
or 24!, T h e n u m b e r of w a y s to arrange 3 0
s o n g s is 30P30, o r 3 0 ! , T h e r e f o r e , t h e t o t a l n u m b e r o f
!)
3 2
s o n g s s o t h a t t h e y a r e t o g e t h e r is ePe • 2 5 . o r 6 ! • 2 5 ,
24P24,
11
I
all p l a y l i s t s .
1320
niO)
=
mp)=220
p l a y l i s t s t h a t c o n t a i n t h e six f a v o u n t e s o n g s t o g e t h e r
N o w determine the probability.
is 2 5
P(C) =
6!
2 4 ! , a n d t h e t o t a l n u m b e r o f p l a y l i s t s is 3 0 ! ,
n(C)
niO)
PiA)
niO)
6!
P(C)
25 24!
PiA)
9
30!
PiC)
6 5 4 3 2 1 V5-24'
PiA)
PiA)
P(A)
a r e l o o n i e s is ^ - ^ , o r a b o u t 0 , 4 0 9 o r 4 0 , 9 % ,
22
30-29-28-27-26
6 5
PiA)
30
4
3
2
22
T h e probability that exactly t w o of t h e d r o p p e d coins
30-29-28-27-26-25-24!
6;5-4-3-2
PiA)
90
220
1
2 8 2 7 26 2 9
L e s s o n 5.4: M u t u a l l y E x c l u s i v e
1
page 338
9 13 2 9
Events,
1. a ) L e t A r e p r e s e n t rolling a s u m o f 2 , L e t B
r e p r e s e n t rolling a s u m o f 8,
1
71 = { 1 , 1), fi = { 4 . 4 } ,
23751
T h e p r o b a b i l i t y t h a t all 6 o f E m a n u e l l a ' s f a v o u n t e
s o n g s will b e p l a y e d t o g e t h e r is
1
or about
23 751
0.000 0 4 2 1 or 0.004 2 1 % ,
F o u n d a t i o n s of Mathemati
>iutions M a n u a l
5-15
c) P(face card or s p a d e ) =
—
P(face card or s p a d e ) =
11
—
T h e p r o b a b i l i t y of d r a w i n g a f a c e c a r d o r s p a d e is
— , or about 0.423 or 4 2 . 3 % .
26
3. a ) Let G r e p r e s e n t g o i n g t o t h e g y m , a n d let S
represent going shopping
b) A a n d B a r e m u t u a l l y e x c l u s i v e , b e c a u s e t h e r e is
no w a y for the s a m e t w o n u m b e r s to a d d up to both 2
a n d 8.
c) Outcome Table:
0.4
O.Ob
Die 1
4
1
2
1
2
3
4
5
2
3
4
5
6
SUM
3
4
5
6
7
4
5
6
7
8
b) N o , t h e s e e v e n t s a r e n o t m u t u a l l y e x c l u s i v e .
P ( G ) + P ( S ) + P ( n e i t h e r ) = 0 . 7 5 + 0.4 + 0.2 = 1.35.
T h i s e x c e e d s 1 , s o t h e r e a r e f a v o u r a b l e o u t c o m e s for
events G and S that are c o m m o n .
T o t a l n u m b e r of o u t c o m e s = 16
P ( s u m of 2 or 8 ) =
2
16
c) P ( G u S) = 1 - P(neither)
1
P ( s u m o f 2 or 8 ) = ^
8
P ( G u S ) = 1 - 0.2
T h e p r o b a b i l i t y t h a t Z a c h will roll a s u m o f 2 o r 8 is
~
0.125 or 1 2 . 5 % .
d) D o u b l e s : { ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) , (4,4)}
S u m o f 6: { ( 2 , 4 ) , ( 3 , 3 ) , (4,2)}
g
P ( d o u b l e s o r s u m of 6 ) =
—
3
P ( d o u b l e s or s u m of 6) = —
T h e p r o b a b i l i t y Z a c h will roll d o u b l e s o r a s u m o f 6 is
- , 0.375 or 3 7 . 5 % .
8
2 . a)
P ( G u S ) = 0.8
T h e p r o b a b i l i t y t h a t M a n a will d o at l e a s t o n e o f t h e s e
a c t i v i t i e s o n S a t u r d a y is 0.8 o r 8 0 % .
4. a) N o . e . g . , 2 is b o t h a n e v e n n u m b e r a n d a p n m e
number.
b) Y e s . e . g . , Y o u c a n n o t roll a s u m of 10 a n d a roll of
7 at t h e s a m e t i m e .
c ) Y e s . e . g . . Y o u c a n n o t w a l k a n d h d e t o s c h o o l at t h e
same time.
5. a)
niA)
PiA)
n(C)
PiA)
=
144 945
389 045
PiA)
28 989
77 809
T h e p r o b a b i l i t y a p e r s o n w h o is M e t i s lives in A l b e r t a
o r British C o l u m b i a is
28989
, or about 0.373 or
77809
10
37.3%.
b) P ( M ) :
niM)
niC)
119 920
PiM)-
389 045
23 984
b) T h e e v e n t s a r e not m u t u a l l y e x c l u s i v e . F o r
e x a m p l e , t h e k i n g o f s p a d e s is b o t h a f a c e c a r d a n d a
spade.
'^^^
77 809
T h e p r o b a b i l i t y t h a t a p e r s o n w h o is M e t i s lives in
M a n i t o b a o r S a s k a t c h e w a n is
23984
or about 0.308
77809
or 3 0 . 8 % .
5-16
C h a p t e r 5: P r o b a b i l i t y
c| Yes, because these two events are mutually
e x c l u s i v e , s o P(A n M) is e q u a l t o 0.
d | T h e o d d s in f a v o u r o f a p e r s o n w h o is M e t i s living
in o n e of t h e f o u r w e s t e r n p r o v i n c e s a r e
2 6 4 8 6 5 ; 124 1 8 0 , or 5 2 9 7 3 ; 2 4 8 3 6 .
i . T h e r e a >• i ' :J..ffed d o g s a n d b e a r s , a n d 2 8 o t h e r
p r i z e s . T h e r e f o r e , t h e o d d s in f a v o u r of w i n n i n g e i t h e r
a s t u f f e d d o g o r a s t u f f e d b e a r a r e 12 : 2 8 , or 3 : 7.
> . 0 bet F r e p r e s e n t t h e d i c e rolls w i t h a s u m o f 5, let
N r e p r e s e n t t h e d i c e rolls w i t h a s u m of 9. Let O
r e p r e s e n t t h e s e t of all d i c e rolls.
R !(:
( 2 , 3 ) . ( 3 . 2 ) . (4.1)}
N: { ( 3 , 6 ) , ( 4 , 5 ) . ( 5 , 4 ) , (6,3)}
n ( F uN) = 8
n ( 0 ) = 36
If 6 8 % o f s e n i o r s h a v e a h e a r i n g l o s s , a n d 1 0 % of
these people do not w e a r glasses, then 1 0 % • 6 8 % , or
6 . 8 % o f s e n i o r s h a v e a h e a r i n g l o s s but d o n o t w e a r
glasses. This m e a n s that 6 1 . 2 % of seniors w e a r
glasses and have a hearing loss,
T h e o d d s a g a i n s t t h e s u m e q u a l i n g 5 or 9 a r e 2 8 : 8,
r(i''
'2',
i ' 2%
! h ^ p: j > j a L i a i > U.at t h i s p e r s o n will not w e a r g l a s s e s
a n d not h a v e h e a n n g l o s s is 1 7 . 2 % .
13. a) Let E r e p r e s e n t t h e e i g h t s , a n d let K r e p r e s e n t
t h e k i n g s . Let O r e p r e s e n t all c a r d s .
or 7 : 2.
b) Let E r e p r e s e n t t h e d i c e rolls in w h i c h b o t h d i c e a r e
e v e n n u m b e r s , a n d let S r e p r e s e n t t h e d i c e rolls w i t h a
s u m of 8. Let O r e p r e s e n t t h e s e t of all d i c e rolls.
E: { ( 2 , 2 ) , ( 2 , 4 ) , ( 2 . 6 ) , ( 4 , 2 ) , ( 4 , 4 ) , ( 4 , 6 ) . ( 6 . 2 ) . ( 6 , 4 ) , (6,6)}
S: { ( 2 , 6 ) , ( 3 , 5 ) , ( 4 . 4 ) . ( 5 , 3 ) , (6,2)}
(2,6), .
a .d n- y) a r e in b o t h s e t s ,
n{F'
>
=
. •
/
'• .
: .)
"••
.•;
/:.',
ct
H)
,
P{G \ H) = 1 4 . 8 %
I h''
'ability t h a t t h i s p e r s o n will w e a r g l a s s e s a n d
n u i ! i.r h e a n n g a i d s is 1 4 . 8 % .
b) Let G r e p r e s e n t w e a r s g l a s s e s a n d H r e p r e s e n t
hnvinn n h - ^ i d n g l o s s .
..',1
n , o % ^ ( 7 6 % + 6.8%)
«(E) = 4
PiEuK)
=
P{EuK)
= 0
.''(/ '
niEuK)
n(0)
/«F; = 4
••nS)-n(EnS)
V
)
'
nit
fi(r _
ni . '
'•/.
=
52
r,2. = o ( E ) + n ( K )
P(EuK) =
13
The o d d s against both dice being e v e n n u m b e r s or
t h e s u m b e i n g 8 a r e 25 : 1 1 .
8. a) Let S r e p r e s e n t s t u d y i n g a n d V r e p r e s e n t p l a y i n g
video games.
= 52
T h e p r o b a b i l i t y o f d r a w i n g a n e i g h t o r a k i n g is ^
P ( S u V) = P ( S ) + PiV) - P ( S n V)
0.8 = 0 4 + 0.6 - P ( S n V)
a b o u t 0 , 1 5 4 or 1 5 . 4 % .
P(SnV)
t h e f a c e c a r d s . L e t O r e p r e s e n t all c a r d s .
= 0.2
b) Let R r e p r e s e n t t h e r e d c a r d s , a n d let F r e p r e s e n t
T o t a l n u m b e r of m e d a l w i n n e r s = 3 0 7 ,
T h e o d d s in f a v o u r o f t h e a t h l e t e h a v i n g w o n t w o o r
m o r e m e d a l s is 6 8 : ( 3 0 7 - 6 8 ) o r 6 8 : 2 3 9 ,
10. e . g , , T n c i a h a s a p r o b a b i l i t y o f 0.3 of c y c l i n g to
s c h o o l o n a n y g i v e n d a y , a n d a p r o b a b i l i t y of 0,2 o f
g e t t i n g a ride f r o m her o l d e r b r o t h e r , S t e v e ,
O t h e r w i s e , s h e w a l k s t o s c h o o l . W h a t is t h e
p r o b a b i l i t y t h a t s h e d o e s not w a l k t o s c h o o l o n a n y
given d a y 2 (0.5)
11. e . g . , T h e r e a r e 6 7 G r a d e 10 s t u d e n t s t h a t t a k e art
a n d 3 7 t h a t t a k e p h o t o g r a p h y . If t h e r e a r e 8 7
students, how m a n y take both2 (17)
12. a) L e t G r e p r e s e n t w e a r i n g g l a s s e s a n d H
represent having a hearing loss.
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
a;
n-'Fi
T h e p r o b a b i l i t y t h a t J o h n will d o b o t h a c t i v i t i e s is 0.2
or 2 0 % ,
b) N o , S i n c e P ( S n V)^0,
t h e n n ( S n V) # 0, s o t h e
s e t s of f a v o u r a b l e o u t c o m e s f o r S a n d V a r e not
disjoint.
9. a) N o , e g . . O n e a t h l e t e w o n t w o o r e m o r e m e d a l s
at t h e S u m m e r a n d W i n t e r O l y m p i c s .
b) T o t a l n u m b e r of m e d a l w i n n e r s = 3 0 7
T h e o d d s in f a v o u r o f a C a n a d i a n m e d a l w i n n e r
w i n n i n g t w o o r m o r e m e d a l s at t h e S u m m e r O l y m p i c s
a r e 2 1 : ( 3 0 7 • 2 1 ) or 21 : 2 8 6 ,
c ) niS u HO = 2 0 + 47 + 1
n ( S uW)
= 6B
, or
r'f
n(.V'
} - 12
/ ) - «.
n(R
>Fj~
ri(R'
t)
h'R)
+ n ( P ) - n(R n
F)
+ 12 - 6
n ( / r ' j f ) ~ 22
frC<i PiR<
F]
t^l
n(RuF)
n{0)
P{R uF)
=
32
52
P{R
^
uP)=
^
13
T h e p r o b a b i l i t y o f d r a w i n g a r e d c a r d o r a f a c e c a r d is
, or a b o u t 0 , 6 1 5 or 6 1 , 5 % ,
13
14. Let D r e p r e s e n t t h e h o u s e h o l d s t h a t h a v e o n e o r
m o r e d o g s , a n d let C r e p r e s e n t t h e h o u s e h o l d s t h a t
h a v e o n e o f m o r e c a t s . L e t O r e p r e s e n t all P r a i h e
households, P(D) = 3 7 % P(C) = 3 1 %
a) P ( D u C ) = 1 0 0 % - P ( ( D u C ) 0
P{Du
C) = 1 0 0 % - - 4 7 %
P ( D u C) = 5 3 %
T h e probability that a Praine household has a cat or
d o g is 5 3 % .
5-17
I^j /-.;;•
' ,
I <i ,\ ~ I'd
i
') , •,
,'./'"'.. t <1 ':.
'
h,",',.
/•(/)
()
!V
in
h t / 'fl I m i l • Pi AI •) + P{AN) + P ( A S P ) + P ( A 8 / V )
' '
I"iI
r { l )
•
f I
i I)
/^(l)anii
24,., t 6 % + 3 % + 1 %
Pptang
147'.
I be pnjbabiliiy ih.d Dani c a n donate blood to t h e next
/-sCj
/'ll'
p e r s o n w i l l ) neo'ds a t r a n s f u s i o n is 4 4 % .
C)
€ ) / - ' i t d r d m i d , - PiABN)
/-({
i'H . • ih
1») '..
I h(> firfib.ihiiitv ifi.'ii .1 i't.IUH- h f i i s r } ! ; ) ' , ! h a s One o r
uitif . r , i ' '
l.nf I.O d ' i ' p
C| P</) •
I
/'(/))
/ •//) • r \
10"-
.''I.'; ,
+ PiAN)
+ P(BW) +
P(Rir;honJ)
1"o + 6 % + 2 % + 7 %
P{Ku;hnnt)-
16'%
P(ON)
1 he p r o b a b i l i t y t h a t R i c h a r d will b e a b l e t o r e c e i v e
b l o o d is 1 6 % ,
'd
19. e.g., T o determine the probability of t w o events
•'.( •„ -
that are not mutually exclusive, y o u must subtract the
.••!/;««•)
probability of both events occurring after adding t h e
i h e [ j t o h . i t i i i i i v th.jt . 1 1 ' f a i r u ; Nou> o i i o i d h a s One o r
m o r e d o g s , o u t n o c a t s , is z z % .
probabilities of each event. Example: Female
students at a high school m a y play hockey or soccer.
1 5 . Let M r e p r e s e n t s n o w o n M o n d a y a n d l e t T
If t h e p r o b a b i l i t y o f a f e m a l e s t u d e n t p l a y i n g s o c c e r is
represent snow o n Tuesday.
6 2 % . t h e p r o b a b i l i t y o f h e r p l a y i n g in g o a l is 4 % . a n d
P(M)
= 60%
t h e p r o b a b i l i t y o f h e r e i t h e r p l a y i n g s o c c e r o r in g o a l is
10%
P(M
6 4 % . t h e n t h e p r o b a b i l i t y o f h e r p l a y i n g in g o a l a t
. / ) - dO%
P{M
\M)
P(M
I)
s o c c e r is 6 2 % + 4 % - 6 4 % = 2 % .
^ P(T)
P{Mn
P)
19. A: s t u d e n t w h o l i k e s r a p m u s i c
(jofo + 4 0 % - 2 0 %
B: s t u d e n t w h o likes b l u e s m u s i c
C: s t u d e n t w h o l i k e s r o c k m u s i c
T h e [)rohol)itity t h a t it w i l l s n o w o n M o n d a y o r o n
O: all s t u d e n t s
T u e s d a y is 8 0 % .
P(A ' itl
16. L e t S r e p r e s e n t d a m a g e t o t h e c o m p u t e r ' s p o w e r
supply a n d let C represent d a m a g e to other
fHA
= 0.15%
(:) = P ( S ) + P(C)
PC ; •
P(.2
u n r of t h e t n r e e g e n r e s ni musir g i v e n ,
- PiS n C)
= 0.15% + 0.30% -
P(0)
0.10%
^ r(C
•(.) = 0.35%
N o o 'j
- 14%
0 ) \ / l ) - 10%
The a s s u m p t i o n is m a d e t h a t 1 0 0 % o f s t u d e n t s like
. C) = 0 . 1 0 %
Pf S- •
13%
' «i) - 20%
r. (1\B)
P((C.
H ( : ) = 0,30%
Pi'>.
<•€))-
I XL ' (A
rnmponents.
R(Sl
C)) ~ 3 0 %
HBHA
Since t h e probability of a n y form of d a m a g e
IS 0 . 3 5 % , t h e c o m p u t e r d o e s n o t n e e d a s u r g e
+ PUB.
P(A , (B
• C ) ) + P { « \ (A u O )
M/i I ' BY) t p ( ( / . , . p ) \ C ) + PUA
1
1 0 0 % - 30"/.. 4 1.3% +- / 0 %
r i>l{A n B)\C)
+ 1 0 % + PiA nBf-^
protector.
17. L e t t h e f o l l o w i n g v a n a b l e s r e p r e s e n t t h e f o l l o w i n g
1 0 0 % = 8 7 % + P[iA
nC)\B)
C)
0 ) \ A) •^ P(A ' ,Dr.
+ 14%
C)
n B)\C)
+ PiAnBn
C)
1 3 % = PUA n B) \ C) + PiA n B n C)
blood types:
OP: type 0 +
T h e p r o b a b i l i t y t h a t a r a n d o m l y s e l e c t e d s t u d e n t will
ON: t y p e
like e i t h e r all t h r e e t y p e s o f m u s i c , o r will like r a p a n d
AP: t y p e A +
b l u e s , b u t n o t r o c k is ' 3 %
AN: t y p e A
2 0 . a ) PiA L. B u C) - P(A) - P ( B ) ^ P ( G )
BP: t y p e B +
b) PiA u B u C) = PiA)
BN: t y p e
P ( / \ r- C) - P ( 8 ^ C l - !^(A
ABP:
type A B +
ABN:
t y p e AB~-
PiB)
4 PfC)
B
P ( A n 6 ) -•
• O)
Applying Problem-Solving Strategies,
a ) i) P ( t y p e O ) = P { O P ) + P { 0 / V )
page 343
P(type O ) = 3 8 % + 7 %
A . e g , , I d o n ' t b e l i e v e t h a t it will c h a n g e t h e
P(type O ) = 4 5 %
The probability that a randomly selected Canadian
p r o b a b i l i t y , b e c a u s e n o t h i n g really c h a n g e s ,
h a s t y p e O b l o o d is 4 5 % .
2
B. There's a - chance of choosing a joke prize, and
o
ii) P ( n e g a t i v e ) = P(OiV) + P(7\/V) + P(BA/) +
PiABN)
P(negative) = 7 % + 6 % + 2%+ 1 %
P(negative) = 1 6 %
The probability that a randomly selected Canadian
a -- chance of choosing the grand prize,
3
h a s a n e g a t i v e b l o o d t y p e is 1 6 % .
Treating t h e j o k e prize a n d the small prize a s
iii) P ( A o r B ) = P(/AP) + P(/\/V) + P(BP) + P ( 8 / V )
equivalent:
P(A or B) = 3 4 % + 6 % + 9 % + 2 %
P ( A o r B) = 5 1 %
T h e probability that a randomly selected Canadian
h a s t y p e A o r B b l o o d is 5 1 % .
5-18
C h a p t e r 5: P r o b a b i l i t y
stay: win grand prize
I'"'"'
3. a ) T h e s e t w o e v e n t s a r e i n d e p e n d e n t ,
b) Let 0 r e p r e s e n t d r a w i n g a d i a m o n d .
switch: win joi<e prize
P(D)= ^
',i:jy vv'in
ptl/t> "
4
joke prize
PiDnD)^
NA'irrh.- w i n g r a n d p r i z e
16
••• s w i t c h : w i n g r a n d p r i z e
After you select, Monty o p e n s a door with a joke
p r i z e . If y o u o r i g i n a l l y s e l e c t e d t h e g r a n d p r i z e , a n d
y o u s w i t c h , y o u v/ill l o s e . B u t if y o u o r i g i n a l l y s e l e c t e d
o n e o f t h e t w o j o k e p r i z e s , a n d y o u s w i t c h , y o u will
w i n . S o , if y o u s w i t c h , y o u will w-n
T h e p r o b a b i l i t y t h a t b o t h c a r d s a r e d i a m o n d s is J L
16
0 0 6 2 5 or 6 . 2 5 % ,
4. a) i) Let 8 r e p r e s e n t L e x i e p u l l i n g a black s o c k f r o m
her d r a w e r ,
PiB)
' <>? ; h e t i m e . T h e
3
b e t t e r s t r a t e g y is t o s w i t c h .
350
1, a i These t w o e v e n t s a r e d e p e n d e n t ,
b) L e t R r e p r e s e n t t h e red die s h o w i n g 4 . let S
r e p r e s e n t rolling a s u m t h a t is g r e a t e r t h a n 7.
=
P'j.^.i,
L ,1 '
ii) Let W r e p r e s e n t L e x i e p u l l i n g a w h i t e s o c k f r o m h e r
drawer
-^^
PiW:
,W)
P{S\R)
Pim
^
t>: = PiB)-i
15
T h e p r o b a b i l i t y o f d r a w i n g t w o b l a c k s o c k s is ^
, or
91
about 0,165 or 16,5%,
l
b
PiR)
Rp;
.
3 5
I n 61 =
'
7 13
15
lnB)==
91
PiW)
P{S -R)
= ^
14
I
L e s s o n 5.5: C o n d i t i o n a l Probability, p a g e
PiR)
j_
4
PiDnD)^
stay: win Joke prize
job
1
4
1
6 2.
P{]A
'/C
PiW)
4
7
PiW
P{W
W)
7
13
}
.
, 14-
12
T h e p r o b a b i l i t y t h a t A u s t i n will w i n a p o i n t
1 >
about 0.0833 or 8.33%,
2 . a) T h e s e t w o e v e n t s a r e d e p e n d e n t ,
b) Let A r e p r e s e n t t h e first c a r d b e i n g a d i a m o n d , a n d
let B r e p r e s e n t t h e s e c o n d c a r d b e i n g a d i a m o n d .
13
PiA)
PiAnB)
52
1
PiA)
PiAnB)
4
12
PiB\A)-
PiAnB)
51
PiB\A)
=
T h e p r o b a b i l i t y of d r a w i n g t w o w h i t e s o c k s is - - , or
i 3
about 0,308 or 3 0 . 8 % .
iii) Let 8 r e p r e s e n t p u l l i n g a b l a c k s o c k , a n d let W
r e p r e s e n t p u l l i n g a w h i t e s o c k . Let A r e p r e s e n t
d r a w i n g a pair o f s o c k s ,
P{A) = PiB nB)+
PiW n W)
1
4
4
17
^
91
13
43
PiA)
91
J_
T h e p r o b a b i l i t y o f d r a w i n g a p a i r o f s o c k s is
17
4
43
- , or
91
about 0,473 or 4 7 . 3 % ,
b) N o , t h e a n s w e r s w o u l d not c h a n g e , b e c a u s e t h e r e
is still no r e p l a c e m e n t .
17
T h e p r o b a b i l i t y t h a t b o t h c a r d s a r e d i a m o n d s is
17
or about 0,0588 or 5,88%,
F o u n d a t i o n s of M a t h e m a t i
.lutions Manual
5-19
5. a) L e t A r e p r e s e n t a s t u d e n t w h o p l a n s t o a t t e n d
U B C , a n d let O r e p r e s e n t all g r a d u a t i n g s t u d e n t s
8. Let S r e p r e s e n t A n i t a r e m e m b e n n g t o set h e r
a l a r m , a n d let N r e p r e s e n t A n i t a n o t r e m e m b e n n g t o
s e t h e r a l a r m . L e t L r e p r e s e n t A n i t a b e i n g late f o r
PiA):
school
n{0)
PfS,
L)
Pp:
1 j -h 0 2 - 0 , 2 0
80
PiS
I )
190
P(N
' ) - PiN)^PiL\N)
P(N
L) - 0 3 8 - 0 . 7 0
PIN
1) - 0 2 t «
30 + 50
80 + 110
PiA)
8
PiA)
19
T h e p r o b a b i l i t y t h a t a g r a d u a t i n g s t u d e n t will a t t e n d
, or a b o u t 0 . 4 2 1 or 4 2 . 1 % .
19
b) Let A r e p r e s e n t a s t u d e n t w h o p l a n s t o a t t e n d
U B C , a n d let P r e p r e s e n t a f e m a l e s t u d e n t .
riS)-PiL\s)
0.124 . 0,266
PiC)
0,39
PiSnL)
124
0
PiC)
U B C IS
PiFA)
=
niP)
niA)
T h e p r o b a b i l i t y A n i t a ' s a l a r m c l o c k w a s s e t is
62
195
or
a b o u t 0 , 3 1 7 or 3 1 . 7 % .
9 Let N r e p r e s e n t a n i c e d a y . a n d let R r e p r e s e n t a
rainy day. Let J represent lan jogging 8 k m ,
PiN n J ) = P(W) • P ( J I N)
P{N n J ) = 0 , 7 0 - 0 , 8 5
PiF\A)J^^
PiN n J ) = 0 , 5 9 5
PiF\A)
=
PiR nJ)
l
= P(R) • P(J
I
R)
PiR n J) = 0 . 3 0 • 0 , 4 0
PiTnC)
= PiT)PiC\T)
P(C)
0.18 ^ 0.56
P ( P n J) = 0 , 1 2
P ( J ) = 0 , 5 9 5 + 0.12
P(J) = 0,715
T h e p r o b a b i l i t y l a n will j o g f o r 8 k m t o m o r r o w is 0 , 7 1 5 .
or 7 1 , 5 % ,
10. Let C r e p r e s e n t a u s e r h a v i n g call d i s p l a y , a n d let
D represent a user having a data plan,
PiTnC)
= 0.30-0,60
P(C)
0,74
PiCnD)^PiD)-PiC\D)
PiTnC)
= 0.18
PiMnC)
P(C
T h e p r o b a b i l i t y t h i s s t u d e n t is f e m a l e is | , 0 . 6 2 5 o r
8
62.5%.
6. Let 7 r e p r e s e n t a t r u e or f a l s e q u e s t i o n , a n d let M
represent a multiple-choice question. Let C represent
a correct question.
PiMnC)
= PiM)
P(MjC)
PiC)
PiC\M)
PiMnC)
=: 0 , 7 0 - 0 . 8 0
PiMnC)
= 0,56
PiM\C)
28
PiD\C)
37
, or a b o u t 0.757 or 7 5 . 7 % ,
37
7. Let F r e p r e s e n t d r a w i n g a l o o n i e t h e first t i m e , a n d
let S r e p r e s e n t d r a w i n g a l o o n i e t h e s e c o n d t i m e .
piPr^S)
^
p ( s | o ^ ^
PiFnS)
'
PiF)
Cj
PiC)
0,30
28
P(F -.S)
0,75
0,30
PiDlC)
T h e p r o b a b i l i t y t h e q u e s t i o n w a s m u l t i p l e - c h o i c e is
p ( o 4
0,40
P{0,
0,56
0,74
PiM\C)
-:D)
PiCnD)
0,70
3
PiDp)
T h e p r o b a b i l i t y t h a t a cell p h o n e u s e r w i t h call d i s p l a y
a l s o h a s a d a t a p l a n is | . or a b o u t 0 , 4 2 9 o r 4 2 , 9 % ,
11. e g , , A s t u d e n t s e l e c t e d at r a n d o m g o e s t o a f a s t f o o d o u t l e t t h a t p a r t i c u l a r d a y . W h a t is t h e p r o b a b i l i t y
that the student had m o r e than 1 h for Iunch2
PiSlP)
= ^ - ^
3 11
=
T h e p r o b a b i l i t y b o t h c o i n s a r e l o o n i e s is
or about
0 , 0 9 1 or 9 , 1 % ,
5-20
C h a p t e r 5: P r o b a b i l i t y
17
ea.
T h e probability that badminton shoes that have lasted
' tc .:;'•:<; 1 , ' , < > - . ; : . ;
, I UivV
<)!l.
ik.y%'
,1 (!
- y..'(
if:. , i
. y.
;-jm
six m o n t h s will last o n e y e a r v. ^
u!/
1< - •
I
.,ii-H.
•
. ;t,.,s|.-f',
\
« til
;«•.
i>-;»v
cycling?
. r a b o u t 0 222 or
22,2%,
16. a ) T h e p r o b a b i l i t y t h a t a m u l t i p l e o f 5 will a l s o b e a
^
H o w o f t e n d o y o u c y c l e w h e n t h e w e a t h e r is
fine?
80
H o w o f t e n d o y o u c y c l e w h e n it is r a i n i n g or
snowing?
60
m u l t i p l e of 3
about 0,286 or 2 8 , 6 % ,
Multiple of 3
Multiple of 5
b) A r a n d o m l y s e l e c t e d s t u d e n t c y c l e d t o s c h o o l o n a
p a r t i c u l a r d a y . W h a t is t h e p r o b a b i l i t y t h a t t h e w e a t h e r
w a s fine that day?
Sf
13. L e t F r e p r e s e n t t i r e s l a s t i n g 5 y e a r s , a n d let S
4-.
r e p r e s e n t t i r e s l a s t i n g 6 y e a r s . S i n c e all t i r e s t h a t
y^'=
,
n
jo
h a v e l a s t e d 6 y e a r s h a v e a l s o l a s t e d 5 y e a r s , it is t r u e
40
•1 *
t h a t P(F n S ) = P ( S ) .
P(S|F) =
P(Fn,S)
,
PiF)
PiS\F)
P(S|F) =
PiS\F)
Ji
=m
PiF)
p,
'
b) Let F r e p r e s e n t a n u m b e r b e i n g a m u l t i p l e o f 5, let
05
7 r e p r e s e n t a n u m b e r b e i n g a m u l t i p l e o f 3.
0.8
F: { 2 0 . 2 5 , 3 0 . 3 5 . 4 0 , 4 5 . 50}
T: { 2 1 , 2 4 , 2 7 , 3 0 , 3 3 . 3 6 , 3 9 . 4 2 , 4 5 . 4 8 }
= 0.625
T h e p r o b a b i l i t y t h a t tires t h a t h a v e l a s t e d 5 y e a r s will
last 6 y e a r s is 0 , 6 2 5 . or 6 2 . 5 % .
14. Let T r e p r e s e n t w i n d s h i e l d w i p e r s l a s t i n g 3 y e a r s ,
a n d let F repres>-n* w i n d s h i e l d w i p e r s l a s t i n g 4 y e a r s .
S i n c e all w i n c n m i r i w i p e r s t h a t h a v e l a s t e d 3 y e a r s
h a v e a l s o la<Tfd 4 ..ears. P ( T n F) = P ( F ) ,
3 0 a n d 4 5 a r e in b o t h s e t s ,
P(T\F) =
PiPnP)
PiF)
2
31
P(T\F) =
T
PiF\T)-
31
P{F)
PiT\F)
' T l
PiF\P)
PiT)
j
T h e p r o b a b i l i t y t h a t a m u l t i p l e of 5 will a l s o b e a
0.6
PiF\T):
PiF\T)
=
=
0.7
m u l t i p l e o f 3 is I , o r a b o u t 0 , 2 8 6 o r 2 8 . 6 % ,
6
17. a) Let X = { 1 s t c h i p is d e f e c t i v e } a n d Y = { 2 n d c h i p
is d e f e c t i v e } .
7
T h e probability that windshield wipers that have lasted
First Chip
Second Chip
3 y e a r s will last 4 y e a r s is - , or a b o u t 0 . 8 5 7 o r
85,7%,
15. Let S r e p r e s e n t b a d m i n t o n s h o e s l a s t i n g six
m o n t h s , a n d let Y r e p r e s e n t b a d m i n t o n s h o e s l a s t i n g
o n e y e a r . S i n c e all b a d m i n t o n s h o e s t h a t h a v e l a s t e d
o n e y e a r h a v e a l s o l a s t e d 6 m o n t h s , PiSnY)
PiYlS)
=
PiY)
PiSnY)
PiS)
p(y|s):
m
PiS)
P(y|s): 0 2
0,9
2
Piy\s)
9
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5-21
b) L e t D r e p r e s e n t a c h i p b e i n g d e f e c t i v e , a n d let N
represent a chip not being defective Let O represent
exactly one of the two chips being defective.
T h e r e are t w o w a y s to d r a w t w o chips, with o n e being
d e f e c t i v e . E i t h e r t h e first is d e f e c t i v e , o r t h e s e c o n d is
defective. Therefore, the probability of both these
events must be d e t e r m i n e d a n d a d d e d together, to
g e t t h e total p r o b a b i l i t y .
PiN)
100
97
P{N\D)
3
PiD\N):
A
150
p^^i^j^
99
PiN n N)
99
291
PiNnD)
PiN)PiD\N)
= 97
^ -_^_^
100 33
97
PiNnD)
=
97
3300
3300
N o w d e t e r m i n e t h e total p r o b a b i l i t y of d r a w i n g e x a c t l y
= P(Or
=
N)
.
3300
>, 1 0 !
=
or about 0,960.
c ) In t h i s c a s e , e i t h e r o n e d e f e c t i v e c h i p is d r a w n first
a n d t h e n o n e n o n - d e f e c t i v e c h i p is d r a w n , o r v i c e
versa. T h e probability of drawing exactly o n e
d e f e c t i v e c h i p is t h e s u m o f t h e p r o b a b i l i t i e s of t h e s e
events
147
97
or about 0.0588 or 5,88%,
1650
e g , . M u l t i p l y t h e p r o b a b i l i t y of t h e first c h i p b e i n g
d e f e c t i v e by t h e s e c o n d c h i p b e i n g n o n - d e f e c t i v e ,
t h e n m u l t i p l y t h e p r o b a b i l i t y of t h e first c h i p b e i n g n o n d e f e c t i v e by t h e s e c o n d c h i p b e i n g d e f e c t i v e , a n d
then add the products,
PiDnN):
P(D)P(A/|D)
PiDnN):
3
150
.
18. L e t D r e p r e s e n t a c h i p b e i n g d e f e c t i v e , a n d let N
r e p r e s e n t a c h i p n o t b e i n g d e f e c t i v e . Let O r e p r e s e n t
e x a c t l y o n e of t h e c h i p s b e i n g d e f e c t i v e ,
a) T h e r e is o n l y o n e w a y to d r a w t w o d e f e c t i v e c h i p s ,
2
^
150
2
PiDp):
= PiD)
149
PiD\D)
3
2
150
149
6
22 350
1
3725
T h e p r o b a b i l i t y of d r a w i n g t w o d e f e c t i v e c h i p s is
1
3725
5-22
149
or about 0,000 268,
PiDnN)
147
149
PiDnN):
441
PiDnN):
•
PiDlN)
147
3
150
149
441
22 350
- P(Dr-,W)
PiDnN)
PiN)
PiDnN)
22""350
P{Dr:N)
150
3
PiDlN)
149
97
PiDnD)
147
PiN)
150
PiN\D)
I
1650
T h e p r o b a b i l i t y of d r a w i n g e x a c t l y o n e d e f e c t i v e c h i p
P(Dr^D)
3577
3725
PiD)
PiO)
P(DoD) -
149
3
Hf
PiDn.D)
146
150
T h e p r o b a b i l i t y o f d r a w i n g t w o n o n - d e f e c t i v e c h i p s is
r . W n D)
97
PiD)
147
22 350
o n e defective chip,
PiO)
PiN\N)
21 462
PiNnN):
3577
33
PiNnD):
9900
PiO)
149
PiN)
PiNnN):
PiNnN)
97
100
=
PiN)
3725
PiDpPiN\D)
PiDnN)
100
99
PiDnN)
chips.
97
3
PiD):
b) T h e r e is o n l y o n e w a y t o d r a w t w o n o n - d e f e c t i v e
PiDnN)
441
441
22 350
22 350
882
22 350
PiD nN)
+PiDnN):
147
3725
T h e probability of drawing exactly o n e defective chip
,
147
or about 0,0395,
3725
19. Let S r e p r e s e n t a s u n n y d a y , a n d R r e p r e s e n t a r a i n y
d a y . Let VV r e p r e s e n t a w i n , a n d L r e p r e s e n t a l o s s ,
a) PiRnW)
= PiR)
PiRnV¥)
PiW\R)
= 0,30 • 0,50
P(R n W) = 0 , 1 5
PiSn
W) = PiS)
•
PiW\S)
PiS n W) = 0,70 • 0 , 6 0
P ( S n W) = 0 , 4 2
PiW)
= PiR n W) + PiS n V¥)
PiW)
= 0,15 + 0,42
PiW)
= 0,57
T h e p r o b a b i l i t y S a v a n n a h ' s t e a m will w i n is 0 , 5 7 . or 5 7 % ,
C h a p t e r 5: Probability
'
i r s t t i m e a n d a total > 7 )
, .4 j r e first t i m e )
a:
o i l i n g a 4 t h e first t i m e )
!
•;
•,
PiS
PiR
Pu
! ,1
: e a m will l o s e is 0 . 4 3 ,
or 4 3 % .
2 0 . e.g.. P r o b l e m 1: O n w e e k d a y s I have cereal for
breakfast 7 0 % o f t h e time. O n the w e e k e n d s I have
cereal for breakfast 4 0 % of the time. O n a r a n d o m
d a y , w h a t is t h e p r o b a b i l i t y t h a t I d o n o t h a v e c e r e a l ?
Solution:
Let WD r e p r e s e n t a w e e k d a y , a n d WE r e p r e s e n t a
w e e k e n d . Let C r e p r e s e n t h a v i n g c e r e a l f o r b r e a k f a s t .
"H'-' ' - t tl -f-^.-sent
not h a v i n g c e r e a l f o r b r e a k f a s t .
' 0 m e a n s P(N | WD) = 0 , 3 0
10 m e a n s PiN | WE) = 0 , 6 0
' Wf )
I WD:
?? ' •' R -1 ••••••'Jnt a c h i p b e i n g d e f e c t i v e , a n d let IV
n - n c " . :,r a - h.p not b e i n g d e f e c t i v e ,
a) I ..I c : , . p , ut t h e s e c o n d c h i p b e i n g d e f e c t i v e ,
••r- • .-I >i -4 t c - lu ,t c h i p d r a w n w a s a l s o d e f e c t i v e . Let
: '.c.-, - ..,.f ib<; (bird c h i p b e i n g d e f e c t i v e , g i v e n t h a t
" ' f < >" L/v. ,vcH a l s o d e f e c t i v e Let O r e p r e s e n t all
'
'inivT'O b e i n g d e f e c t i v e . In t h i s c a s e , t h e r e
- o ' . y ,u>- vvav -o d r a w t h r e e d e f e c t i v e c h i p s .
p(r) =
98
1
PtP) = - 2
^ '
49
p ( 0 ) 4
PiO) =
PiO)
PiD)-PiS)-P(T)
JL JL JL
25
p ( s ) 4
33
49
1
P(0) =
40425
NI WE
T h e p r o b a b i l i t y o f d r a w i n g 3 d e f e c t i v e c h i p s is
Cl.
J
42
ro
;
70
40 425
or a b o u t 0,000 024 7 or 0.002 4 7 % ,
b) Let B represent the s e c o n d chip not being
d e f e c t i v e , g i v e n t h a t t h e first c h i p d r a w n w a s a l s o not
d e f e c t i v e . L e t C r e p r e s e n t t h e third c h i p n o t b e i n g
d e f e c t i v e , g i v e n t h a t t h e first t w o w e r e a l s o not
d e f e c t i v e . Let F r e p r e s e n t all t h r e e c h i p s d r a w n not
b e i n g d e f e c t i v e . In t h i s c a s e , t h e r e is o n l y o n e w a y t o
draw three non-defective chips.
](•>',(>
100(1
PiN):
4900
1890
4900
PiN):
27
96
70
PiN)
Problem 2: I draw two cards from a well shuffled
standard deck, drawing the second card without
r e p l a c i n g t h e first o n e . If m y s e c o n d c a r d is a red
c a r d , w h a t is t h e p r o b a b i l i t y t h a t m y first c a r d is b l a c k ?
S o l u t i o n : Let B r e p r e s e n t t h e first c a r d b e i n g b l a c k
a n d R r e p r e s e n t t h e s e c o n d c a r d b e i n g red W e a r e
l o o k i n g f o r P{B | R).
5 2 ^
26
26
26
or a b o u t 0 , 5 1 0
51
2 1 . e g , . T h e p r o b a b i l i t y of e v e n t A a n d b b o t h
o c c u r n n g is t h e p r o b a b i l i t y A o c c u r s , m u l t i p l i e d by t h e
p r o b a b i l i t y B o c c u r s g i v e n t h a t A o c c u r s . E x a m p l e : If a
6 - s i d e d d i e is r o l l e d t w i c e ,
Foundations of Mathemati
98
24
P(C) =
^
^ '
49
25
95
47
25'99
24
49
107160
P(F)
=
121275
8085
T h e p r o b a b i l i t y o f d r a w i n g 3 d e f e c t i v e c h i p s is - 1 — ,
8085
or a b o u t 0 , 8 8 4 o r 8 8 , 4 % ,
51
Answer:
P(C)
'
P{N)P{B)P{C)
PiF)
PiF)
26
52
P{B\R
99
7144
^IHrT
P(B|R)=
=
PiF)
PiBnR)
26
95
PiB)
100
27
Answer: — or about 0,386
70
Pih\R]
1
.lutions Manual
c ) In t h i s c a s e , m o r e d e f e c t i v e c h i p s will b e p u l l e d
w h e n 2 or 3 d e f e c t i v e c h i p s a r e p u l l e d . T h e r e a r e
3 d i f f e r e n t w a y s t o pull 2 d e f e c t i v e c h i p s .
Let S represent the s e c o n d chip being defective,
g i v e n t h a t t h e first c h i p d r a w n w a s a l s o d e f e c t i v e . Let
U r e p r e s e n t t h e third c h i p n o t b e i n g d e f e c t i v e , g i v e n
t h a t t h e first t w o w e r e d e f e c t i v e . Let G r e p r e s e n t t w o
o f t h e t h r e e c h i p s d r a w n b e i n g d e f e c t i v e . Let A
r e p r e s e n t at l e a s t t w o of t h e t h r e e c h i p s d r a w n b e i n g
defective.
5-23
_4
99
100
P(S) = 1
33
Jl
25
PiD):
T h e p r o b a b i l i t y C e l e s t e will u s e a s t a t i o n a r y b i k e a n d
3
P(S) =
^ ^
98
Piuy='^
^ '
49
f r e e w e i g h t s f o r h e r n e x t w o r k o u t is | , o r a b o u t 0 . 1 6 7
6
or
m:!%.
P{G)
:P(D)-P(S)-P(ti)-3
3. a) T h e t w o w o r k o u t s a r e d e p e n d e n t . A f t e r l a n
c h o o s e s a workout, he won't c h o o s e the s a m e o n e
P{G)
±
±
3
^ 25 ' 33 ' 49 ~
again.
b) L e t T r e p r e s e n t l a n r u n n i n g t h e t r a c k , a n d let E
r e p r e s e n t l a n u s i n g a n elliptical w a l k e r .
144
PiG)
40425
'V,
48
P{G)
13475
pp
/ >
Pi 7
F)
;
P2' )
T h e total p r o b a b i l i t y c a n n o w b e d e t e r m i n e d .
PiA)
PiG)
t P(0)
48
P(^)
1
I
=
P'E\T)
1
Pi I : F ;
1/
13475 ' 40425
144
PiA)
p(£|nP
^ _ 1
T h t i pn.r),ic-!iiiy t h a t l a n will r u n t h e t r a c k a n d u s e t h e
40425 ' 40425
elliptical w a l k e . 's
^
or a b o u t 0 0 8 3 3 o r 8 . 3 3 % ,
145
PiA)
4. a | I fn /V . e p i e s o M i : p i n n i n g a r e d , a n d let T
r e p ' O : ' I '• . . T ' o o '! ( w o .
1
40425
29
PiA)
8085
T h e probability of drawing m o r e defective chips than
n o n - d e f e c t i v e c h i p s is
29
, or a b o u t 0.0036 or
8085
P(R.
J)
P{R.
r(R)
PiT)
'I
'
4
6
I)
0.36%.
PiP
JL
/ )
L e s s o n 5.6: I n d e p e n d e n t E v e n t s , p a g e 360
M
T h e p r o b a b i l i t y t h e s p i n n e r will l a n d o n r e d a n d t h e d i e
1. a) T h e s e e v e n t s a r e i n d e p e n d e n t , b e c a u s e t h e
result of t h e s p i n n e r d o e s n o t a f f e c t t h e result of t h e
die, and vice versa.
Will l a n d o n 2 is ^ , o r a b o u t 0 . 0 4 1 7 o r 4 . 1 7 % .
24
b) T h e s e e v e n t s a r e i n d e p e n d e n t , b e c a u s e t h e result
of t h e red d i e d o e s n o t a f f e c t t h e result of t h e g r e e n
die, a n d vice versa.
c ) T h e s e e v e n t s a r e d e p e n d e n t . B e c a u s e t h e r e is n o
r e p l a c e m e n t , t h e d e c k t h a t t h e s e c o n d c a r d is d r a w n
f r o m is t e c h n i c a l l y d i f f e r e n t t h a n t h e o r i g i n a l d e c k .
d) T h e s e e v e n t s a r e i n d e p e n d e n t , b e c a u s e
r e p l a c e m e n t is o c c u r n n g . w h i c h r e s e t s ' t h e p r o b a b i l i t y
for e a c h draw.
2. a) T h e s e e v e n t s a r e likely i n d e p e n d e n t . T h e c a r d i o
workouts focus on the heart, and use the legs the
most often. Therefore, there w o u l d be no specific
reason w h y o n e w o r k o u t w o u l d be favoured over
another.
b) L e t O r e p r e s e n t rolling a o n e o n t h e r e d d i e , a n d F
r e p r e s e n t rolling a f i v e o n t h e g r e e n d i e .
1
PiO)
PiOnF)
:-PiO)
PiOnF)
=
l l
PiOnF)
=
-^
P{F)
T h e p r o b a b i l i t y of rolling a 1 o n t h e r e d d i e a n d a 5 o n
t h e g r e e n die is 4 : - or a b o u t 0 . 0 2 7 8 o r 2 . 7 8 % .
36
b) Let B r e p r e s e n t C e l e s t e u s i n g a s t a t i o n a r y b i k e ,
a n d let F r e p r e s e n t C e l e s t e u s i n g f r e e w e i g h t s .
PiB) =
PiB nP)
P(enP)
PiBnP)
5-24
l
PiF)
- PiB)
\
PiF)
1
1
3
2
1
: ^
C h a p t e r 5: P r o b a b i l i t y
c ) Let K r e p r e s e n t d r a w i n g a k i n g , a n d let A r e p r e s e n t
drawing an ace.
6. a)
First Child
Second Child
Pi
Pi'boy) = 0,5^
Pi
P(boy) = 0,5
PiA
51
A\K)
P{
,P(girl) = 0.5
Th'
L.a-'.tv
.g will b e d r a w n first, a n d a n a c e
P{girl| = 0.5
will b e d r a w n s e c o n d is —
, or about 0.006 or 0,6%.
663
d) Let N r e p r e s e n t d r a w i n g a p r i m e n u m b e r , a n d let F
r e p r e s e n t d r a w i n g a m u l t i p l e of five.
10
PiN)
P(F) = 3 ±
^ 30
30
PiB,
PIP,
P'P,
P < . , P
P{NnF)
=
PiN)PiF)
1 1
P{NnF)
3
P{NnF)
5
1
=
b) T h e g e n d e r s o f t h e c h i l d r e n a r e i n d e p e n d e n t
e v e n t s . A l s o , t h e r e is o n l y o n e w a y t h a t b o t h c h i l d r e n
can h ^ boyc Let B represent a child being a boy.
15
T h e p r o b a b i l i t y t h a t t h e first c a r d d r a w n is p n m e a n d
Pp'O
F)
Fj
E} =
'• 5
\ VB)
PiB)
1.5 0,5
0 25
1 he p r o b a b i l i t y t h a t b o t h c h i l d r e n a r e b o y s is 0 , 2 5 . o r
25%.
c) T h e r e are t w o that o n e child can be a boy a n d the
o t h e r c a n b e a giri.
Let B r e p r e s e n t a c h i l d b e i n g a b o y , a n d let G
r e p r e s e n t a c h i l d b c m g a giri.
PtP) " 0 f
p(c;) - 0 5
t h e s e c o n d c a r d is a m u l t i p l e of 5 is J L , o r a b o u t
15
FiH,
0,0667 or 6 . 6 7 % .
r ( B . . -2) - o c
5. a) If t w o e v e n t s a r e i n d e p e n d e n t , t h e n P(A n B) is
f i : - jHohnbi.iiv , n ^ ! <.ne c h i l d is a b o y a n d t h e o t h e r is
a giri is 0 5. o r 5 0 % .
e q u a l t o PiA)
PiA nB)
• P(8).
= 0,12
PiA)
• PiB)
= 0 3 5 • 0,4
PiA)
• P(B) = 0,14
S i n c e PiA nB)*
P(A) • PiB),
n o . t h e s e e v e n t s a r e not
independent,
b) PiA nB)
= 0 468
PiA)
• PiB)
= 0,720 • 0,650
PiA)
• PiB)
= 0 468
S i n c e PiA n B) = PiA)
• PiB),
yes. these events are
^ n, - p{B]
PiB , , 'P
- 0 '•
Pi(-)
fj 5
•2
2
7. T h e s e e v e n t s a r e d e p e n d e n t , b e c a u s e t h e d e c k
h a s 4 0 c a r d s w h e n t h e first c a r d is d e a l t , b u t it h a s
3 9 c a r d s w h e n t h e s e c o n d c a r d is d e a l t .
Let C r e p r e s e n t a c l u b b e i n g d e a l t , a n d let H
represent a heart being dealt.
PiC)
^°
40
PiC)
1
4
PiCnH)
10
P(Cr^P)
independent.
P(P|C) =
PiCnH)
PiC)P{HlC)
1
10
4
39
10
156
39
PiCnH)--
5
78
T h e p r o b a b i l i t y t h e first c a r d d e a l t is a c l u b a n d t h e s e c o n d
c a r d d e a l t is a h e a r t is — , or a b o u t 0 . 0 6 4 1 or 6 . 4 1 % .
78
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5-25
S. L e t A r e p r e s e n t t h e first roll. ar. i /-
t resent the
second.
a) I ' l 4
M
id/T
fdV^l
ip
'
^
c ) T h e p r o b a b i l i t y J e r e m i a h will s e e o n l y o n e o f t h e s e
a n i m a l s is t h e s a m e a s t h e p r o b a b i l i t y t h a t h e will n o t
see both or neither of t h e m . So. subtract these
probabilities f r o m 1 to determine the n e w probability.
Let O r e p r e s e n t J e r e m i a h s e e i n g o n l y o n e o f t h e s e
animals.
PiO)
A
PiCcB)-
Pii.
p.
.<h
T h e p r o b a b i l i t y is
36
| j | .'HA
T^i
/'i/i;
o
P(0) =
^
^ '
20
T h e p r o b a b i l i t y J e r e m i a h will s e e o n l y o n e o f t h e s e
l":l<]
a n i m a l s is
. 0.35 or 3 5 % .
20
1 0 . a) T h e s e t w o e v e n t s a r e i n d e p e n d e n t . L e t H
r e p r e s e n t g e t t i n g h e a d s f r o m t o s s i n g a c o m . a n d let S
r e p r e s e n t s p i n n i n g a six o n a s p i n n e r .
r;
P(AnB):
PiH)
PiS)
PiHnS)
1 he p t o b . i b i l i l v 1'^
2
c ) P(A
B] - l-'lAi
P'lil
^ '
12
' /V^)
2
' •'i2
PiS)
^ '
^
6
T h e p r o b a b i l i t y o f s p i n n i n g a six is
T h e p r o b a b i l i t y is
| | .
9. T h e s e t w o e v e n t s a r e i n d e p e n d e n t . L e t C r e p r e s e n t
s e e i n g a c a m e l , a n d let B r e p r e s e n t s e e i n g a n ibis.
a) P{CnB)r
P{C)
P(CnB) =
PiCnB)
l
P{B)
|
I
T h e p r o b a b i l i t y J e r e m i a h will s e e a c a m e l a n d a n ibis
is ? , 0 6 o r 6 0 %
5
b) P ( C ' )
1
P(C)
PiB')
^-P{B)
1
PiB')
1-^
P(C')
^
P(S')
I
P(C3'oB')
PiC)
PiB')
=
20
T h e p r o b a b i l i t y J e r e m i a h will s e e n e i t h e r a n i m a l is
1
, 0.05 or 5%.
20
P ( S ) 4
T h e p r o b a b i l i t y o f s p i n n i n g a six is
PiC)
PiC'nB')
| .
6
e g.. S p i n n e r h a s 6 e q u a l a r e a s , n u m b e r e d 1 t o 6.
b) T h e s e t w o e v e n t s a r e i n d e p e n d e n t . Let H
r e p r e s e n t g e t t i n g h e a d s f r o m t o s s i n g a c o i n , a n d let S
r e p r e s e n t s p i n n i n g a six o n a s p i n n e r ,
PiH)-PiS)
=
PiHnS)
.
e . g . , S p i n n e r h a s 10 e q u a l a r e a s , n u m b e r e d 1 to 10,
11. a) A n n e ' s first b a g c o n t a i n s 7 r e d m a r b l e s a n d
3 b l u e m a r b l e s , a n d h e r s e c o n d b a g c o n t a i n s 4 red
m a r b l e s a n d 5 b l u e m a r b l e s . B e c a u s e s h e is d r a w i n g
2 marbles from 2 separate bags, the two events are
independent.
Let F r e p r e s e n t a b l u e m a r b l e d r a w n f r o m t h e first
b a g . a n d let S r e p r e s e n t a b l u e m a r b l e d r a w n f r o m t h e
s e c o n d b a g . Let B r e p r e s e n t d r a w i n g t w o b l u e
marbles,
P(P) ^ '
10
PiB) =
P(S) =
|
9
PiF)-PiS)
3 5
P(e)=2P,P
10
9
P(B)P
5-26
C h a p t e r 5: P r o b a b i l i t y
T h e p r o b a b i l i t y A n n e will d r a w t w o b l u e m a r b l e s is 1
6 '
or a b o u t 0 . 1 6 7 o r 1 6 . 7 % ,
Abby i - ! ' ' ; .
re pre - f n i i. a v ;
drawr••; >
J • o =.;,|.' .
t 1
A b b y ' s b a g c o n t a i n s 11 red m a r b l e s a n d 8 b l u e
m a r b l e s . S i n c e s h e is d r a w i n g 2 m a r b l e s f r o m t h e
s a m e bag, the two events are dependent.
/ ••( 1)A,
iA 1 /
Fll
P(F}
Let F r e p r e s e n t t h e first m a r b l e d r a w n b e i n g b l u e , a n d
let S r e p r e s e n t t h e s e c o n d m a r b l e d r a w n b e i n g b l u e .
P(e|o,.-;;
P{OA)
Let B r e p r e s e n t d r a w i n g t w o b l u e m a r b l e s
P{B|.c':
/n')/».)
P(F)
=
P(S)-=P(F-yPmF)
19
P(B):
P(S F) = A
^
'
18
8_
^
19
18
ig a r e d m a r b l e , a n d let B
rble. Let OA r e p r e s e n t
one blue marble.
/ \ll
'
R)
Let A r e p r e s e n t t h a t A n n e a n d A b b y w i l l b o t h d r a w
o n e red m a r b l e a n d o n e b l u e m a r b l e .
F[A\
>'{(:•]
R{nA)
\I\
T h e p r o b a b i l i t y A b b y will d r a w t w o b l u e m a r b l e s is
d
n l 0_
IJ]
;'.
^ P ^ , o r a b o u t 0 . 1 6 4 or 1 6 . 4 % .
F>A,
No. e.g., A n n e has a
or
171
a b o u t 0 . 1 6 4 p r o b a b i l i t y of d r a w i n g t w o b l u e m a r b l e s ,
b) F o r b o t h g i d s , t h e r e a r e t w o w a y s t h a t t h e y c a n
d r a w o n e red m a r b l e a n d o n e b l u e m a r b l e . H o w e v e r ,
in A b b y ' s c a s e , t h e p r o b a b i l i t i e s will b e t h e s a m e ,
because the events are dependent. Therefore, only
o n e probability needs to be d e t e r m i n e d , and that
result c a n b e m u l t i p l i e d b y 2 t o d e t e r m i n e t h e total
probability.
A n n e : Let F r e p r e s e n t A n n e d r a w i n g a red m a r b l e
f r o m her first b a g . a n d a b l u e m a r b l e f r o m h e r s e c o n d
b a g . a n d let S r e p r e s e n t A n n e d r a w i n g a b l u e m a r b l e
f r o m her first b a g a n d a red m a r b l e f r o m h e r s e c o n d
b a g . Let O r e p r e s e n t H-n.^inn o ^ e red m a r b l e a n d
o n e blue marble.
P:7)
'
10
Pi
f
.'395
- or a b o u t 0 . 1 6 7 p r o b a b i l i t y of
6
drawing two blue marbles, while A b b y has a
,
^
^
9
v.r
•068
:y b o t h giris will e a c h d r a w o n e r e d a n d
o n e b l u e m a r b l e is
, or about 0 2 6 9 or 2 6 . 9 % .
7695
c ) A n n e : L e t F r e p r e s e n t d r a w i n g a red m a r b l e f r o m
t h e first b a g , a n d let S r e p r e s e n t d r a w i n g a red m a r b l e
f r o m t h e s e c o n d b a g . Let R r e p r e s e n t d r a w i n g t w o red
marbles.
P{F)^^
10
PiR)
P{F)
P(P)^11
1
2
PiS)
P{S)
l
PiR)
2
2
T h e p r o b a b i l i t y A n n e will d r a w t w o red m a r b l e s is
| ,
0 25 or 2 5 % .
A b b y : Let F r e p r e s e n t t h e first m a r b l e d r a w n b e i n g
r e d , a n d let S r e p r e s e n t t h e s e c o n d m a r b l e d r a w n
b e i n g r e d . Let R r e p r e s e n t d r a w i n g t w o r e d m a r b l e s .
P ( 0 ^ 4
P(0)
Pi:,
i
= P(F) + P(S)
P ( 0 ) ^ U ^
18
15
PiP)
10
-i,-^
20
PiO)
1
PiO)
PiF)P(S|F) = |
90
90
P(0) = ^
^ '
PiO)
. P{F)
=
P{S\F)
1
9
2
19
38
T h e p r o b a b i l i t y t h a t A b b y d r a w s t w o r e d m a r b l e s is
90
, o r a b o u t 0 , 2 3 7 or 2 3 , 7 % ,
38
N o , e g , , A n n e h a s a ~~ or 0 , 2 5 p r o b a b i l i t y of d r a w i n g
t w o red marbles, while A b b y has a — or about 0,237
38
p r o b a b i l i t y o f d r a w i n g t w o red m a r b l e s .
F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5-27
12. T h e t w o drawings are independent, b e c a u s e the
first t r e a t is r e p l a c e d a f t e r it is p i c k e d .
L e t G r e p r e s e n t d r a w i n g a g r a n o l a bar, let F r e p r e s e n t
d r a w i n g a fruit b a r , a n d let C r e p r e s e n t d r a w i n g a
c h e e s e strip.
a ) T h e s e c o n d g r a n o l a b a r c a n b e k e p t if a g r a n o l a
b a r is d r a w n b o t h t i m e s .
P(6) = I I
PiGnG)
P{G)
b) PiW')
1
PiW')
1
P(14^)
P(W'nXp=-PiW')-P(X')
1
P(W'nX')
'
20
19
PiW)
•
PiW'r,X')
^ - ^
20 20
361
400
20
PiW')
=
'
PiX')
361
T h e p r o b a b i l i t y t h a t T i e g a n w i n s n e i t h e r p n z e is
P{G)
400
0,9025 or 9 0 , 2 5 % ,
T h e p r o b a b i l i t y t h a t a g r a n o l a b a r is k e p t is | , 0 . 2 5 o r
25%.
b) I r-t .K r ) - | . ( o s r n i •^o-'-pinc] a treat.
A n y tft ai - . . h i b e k e p t if i w o of t h e s a m e t r e a t s a r e
ftr-.iwn
14. a) e . g . . P r o b l e m : W h a t is t h e p r o b a b i l i t y o f
drawing a card from a shuffled standard deck and
g e t t i n g a red c a r d , t h e n r e p l a c i n g it, s h u f f l i n g t h e d e c k
again, drawing a second card, and getting a heart?
S o l u t i o n : Let R r e p r e s e n t g e t t i n g a r e d c a r d a n d H
reptesf>nt g e t t i n g a h e a d
P|
P(P
op
P{R'
P)
FiF
20
F(F
Fl
/ ' ( , ' - ) P;
F(F,
,F)
-
4
^ ^
200
- P{C)
PiC)
20
= PiG nG)
^ ^
P(C o C )
,/p
P{R<
,H)
400
79
,
200
0 . 3 9 5 or 3 9 . 5 % .
= 1
PiK)
= 121
200
P(Pi
26
P{3\R]
79_
» i P , S|
P|P) P(S|P)
200
,
P P. , 5 |
^
'
26
—
52
, 0.605
13
13. B e c a u s e t h e w i n n e r ' s t i c k e t is r e t u r n e d t o t h e
d r a w a f t e r t h e first p r i z e is a w a r d e d , t h e t w o e v e n t s
are independent.
Let W r e p r e s e n t T i e g a n w i n n i n g t h e first p r i z e , a n d let
X represent winning the s e c o n d prize.
a) P(14^) = ^
100
PiWnX)
. P(U1/) P ( X )
-
51
13
P(PnS) = 02
T h e p r o b a b i l i t y is
or 6 0 . 5 % .
P{X) -
1
' . 0 125 o r 1 2 . 5 % .
8
02
T h e p r o b a b i l i t y t h a t a t r e a t is not k e p t is
PiW)
"i
b) e . g . . P r o b l e m : W h a t is t h e p r o b a b i l i t y of d r a w i n g a
c a r d f r o m a s h u f f l e d s t a n d a r d d e c k a n d g e t t i n g a red
card, then drawing a second card without replacing
t h e first o n e , a n d g e t t i n g a s p a d e ?
S o l u t i o n : Let R r e p r e s e n t g e t t i n g a r e d c a r d o n t h e
first d r a w , a n d S r e p r e s e n t g e t t i n g a s p a d e o n t h e
second draw.
= 1 •• P { K )
Pin
'V
1
4
1 ho p r o b a n i l i t y is
+ PiF n F ) + P ( C n C )
400
52
1
2
'^[H]
I 'i
«
T h e p r o b a b i l i t y t h a t a n y t r e a t c a n b e k e p t is
c ) PiK)
r(p.
/
20
PiK)
1
P(pi
13
, or a b o u t 0,127 or 12,7%,
15. T h e r e a r e n i n e w a y s t h a t t w o s i n g l e - d i g i t n u m b e r s c a n
a d d up to 10. and there are 100 w a y s to select t w o singledigit n u m b e r s . T h e r e f o r e , t h e p r o b a b i l i t y t h e s u m o f t h e s e
t w o n u m b e r s is 10 is ^ .
100
0,09 or 9 % ,
^
20
20
T h e p r o b a b i l i t y T i e g a n w i n s b o t h p n z e s is
400
0,0025 or 0 , 2 5 % ,
5-28
C h a p t e r 5: P r o b a b i l i t y
16
wh-
' ^
, /•
•
' i PiA n B) = PiA) ^ PiB) only
idependent events.
red m a r b l e s f r o m a b a g
."•c
r,Me m a r b l e s , w i t h
id o n 1st d r a w } a n d
m} a r e i n d e p e n d e n t e v e n t s , s o
b)
COi
B =
Pi
fi)
18. Let W r e p r e s e n t a w i n , a n d let L r e p r e s e n t a loss
a) T h e r e is o n l y o n e w a y that f o u r g a m e s in a r o w c a n
Let F r e p r e s e n t w i n n i n g t h e s e n e s in 4
games.
•'.p.
I iW)
- PiW).
PiW)
-
PiW)
c.'.65f
F.I ] - o 1 7 8 . . .
: h. ..r .[.ability t h a t M o n t r e a l w o u l d w i n t h e s e r i e s in
IUU. y c . i . e s is a b o u t 0 . 1 7 9 o r 1 7 . 9 % .
b) S i n c e t h e o d d s in f a v o u r of M o n t r e a l w i n n i n g f o u r
g a m e s in a r o w w e r e 5 : 4 , t h e p r o b a b i l i t y of t h i s
P(
wo red m a r b l e s f r o m a b a g
••S'^ I S h'ue m a r b l e s , w i t h o u t
1 t draw} and
c)
COI
ref
pendent events so
B = {r
occurring w a s |
•A'- ,.n i ifu • .>:
Fif
}
I ii..
, ".I ,
P'vV]
I
/ HV,
.-V
o :( ^
o r a b o u t 0 . 5 5 5 . . . . Let P r e p r e s e n t
in 4 g a m e s .
- VA) • P(W)
-
PiW)
P{A)-PiBlA)
Pi
P(
ti root t w i c e ;
i he p:_ibub,.iV M o n t r e a l w o u l d w i n a n y s i n g l e g a m e in
the s e n e s w a s about 0.863. . or 8 6 , 3 % , a s s u m i n g the
c h a n c e of winning any of the four g a m e s w a s equal.
17 ' r - i nl I.
ch part is i n d e p e n d e n t . L e t F
n g , a n d let O r e p r e s e n t n o p a r t s
rep
failiogK
a) PiP) = 0-01
P ( F 1 = 0.99
B e c a u s e e a c h e v e n t is i n d e p e n d e n t , t h e p r o b a b i l i t y
t h a t n o p a r t s fail is d e t e r m i n e d by m u l t i p l y i n g t h e
p r o b a b i l i t y t h a t o n e part d o e s n o t fail by itself
100 times,
PiO) =
iPiR)r'
PiO) =
iOMf°
19. T o p a s s e a c h test, G a v i n m u s t a n s w e r at l e a s t
t h r e e q u e s t i o n s c o r r e c t l y . For t h e t r u e - f a l s e test, t h e
p r o b a b i l i t y o f a n s w e n n g e a c h q u e s t i o n c o r r e c t l y ( C ) , is
0,5, a s is t h e p r o b a b i l i t y of a n s w e n n g it w r o n g l y ( W ) , B y
the tree diagram, there are 32 different possible '
o u t c o m e s , each one with an equal chance of occurring
( 0 . 0 3 1 2 5 ) , Of t h e s e 3 2 o u t c o m e s , t h e r e a r e 16 in w h i c h
at least 3 a n s w e r s are correct so the probability of
p a s s i n g t h e t r u e - f a l s e t e s t is, | |
w a y t o c a l c u l a t e t h i s is 16 • 0 , 0 3 1 2 5 = 0,5,
P ( 0 ) = : 0.366...
T h e p r o b a b i l i t y t h a t t h e m a c h i n e will o p e r a t e
c o n t i n u o u s l y f o r o n e y e a r is a b o u t 0 . 3 6 6 or 3 6 . 6 % .
b) PiF) = 0 . 0 0 5
P ( P ) = 0.995
D e t e r m i n e t h e p r o b a b i l i t y t h a t n o p a r t s c a n fail in t h e
same way.
PiO) =
iPiF')f"'
P(0) = (0.995f°
W i t h t h e m u l t i p l e - c h o i c e test, t h e p r o b a b i l i t y of a n s w e n n g
a q u e s t i o n c o r r e c t l y is ^
o u t c o m e s . 16 o f w h i c h a l l o w f o r at least 3 c o r r e c t
a n s w e r s . T h e s e o u t c o m e s a r e not e q u a l l y likely.
One outcome has 5 correct answers,
f
P(5 correct, 0 wrong) -
T h e p r o b a b i l i t y t h e m a c h i n e will o p e r a t e c o n t i n u o u s l y
f o r o n e y e a r is a b o u t 0 . 6 0 6 o r 6 0 . 6 % .
PiO) =
Five o u t c o m e s have 4 correct a n s w e r s a n d one w r o n g
(i)'
• —
y4 j
y4y
Ten outcomes have 3 correct answers and 2 wrong
0,998...
T h e p r o b a b i l i t y of not failing n e e d s t o b e a b o u t 0 9 9 9
o r 9 9 . 9 % t o e n s u r e that t h e p r o b a b i l i t y t h a t t h e
m a c h i n e will o p e r a t e c o n t i n u o u s l y is 9 0 % .
F o u n d a t i o n s of M a t h e m a t i
(3)
answer, 5 • P(4 correct, 1 wrong) = 5
'io^^'iiPinr
=
1
-~
iPiF')f''
0.9=(P(P)r
PiP')
and the probability of
a n s w e r i n g it w r o n g l y is | . T h e r e a r e a l s o 3 2 p o s s i b l e
P ( 0 ) = 0.605...
c)
o r 0,5. N o t e : A n o t h e r
-lutions M a n u a l
a n s w e r s , 10
P(3 correct, 1 wrong) = 1 0
r i f
-
•
f s f
^
5-29
M a t h in A c t i o n , p a g e 3 6 3
P l a t least 3 correct a n s w e r s
T h e group found that the e x p e n m e n t of drawing o n e
ball at a time w a s the most productive. T h i s is
b e c a u s e w h e n two balls are drawn at o n c e ,
conditional probability exists, w h i c h c a n be hard to
deal with.
+ 1014
1
15
1024
1024
1024
106
1024
53
or 0.1035...
512
Ttie probability of p a s s i n g the multiple-choice test is
53
or about 0 . 1 0 3 5 .
512
T r e e for true-false test
T h e best w a y to m a k e the conjecture more reliable is
to i n c r e a s e the number of tnals. W h e n this is d o n e ,
the e x p e n m e n t a l probability slowly begins to act more
like the theoretical probability.
H o w e v e r , there is no type of e x p e n m e n t that c a n
g u a r a n t e e a correct conjecture. T h i s is b e c a u s e
theoretical probability is just that; theoretical. T h i s
probability is only g u a r a n t e e d to work after millions or
possibly billions of tnals, or continuous tnals.
T r e e for multiple-choice test
" C 0.25
Chapter Self-Test, page
364
C0.2S
1. O u t c o m e T a b l e
I
Tile 1
2
SUM
I
4
\i
2
3
4
5
^•111
6
CM
2
3
4
5
6
7
i>
3
4
5
6
7
8
9
4
5
6
7
8
9
10
6
7
8
9
10
11
7
8
9
10
11
6
P odd s u m =
18
P(even sum):
1
8
12
18
36
36
P(even sum):
2
2
Y e s , e.g., T h e probabilities of winning are e q u a l .
Plodd sum
2 . Let A represent a n Inuit p e r s o n being able to
c o n v e r s e in at least two Abonginal l a n g u a g e s .
P{A) =
^
7+ 3
P{A)
10
T h e probability that a n Inuit person c a n c o n v e r s e in at
least two Abonginal l a n g u a g e s is
5-30
, 0.7 or 7 0 % .
C h a p t e r 5: P r o b a b i l i t y
3. L e t A represent a p a s s c o d e with 3 d i f f e r e n t e v e n
niOp
d i g i t s , a n d l e t O r e p r e s e n t a l l 3^digit p a s s c o d e s .
T h e r e a r e 5 d i f f e r e n t p o s s i b i l i t i e s f o r e v e n d i g i t s : 0,
52!
nipp
2,4,6, a n d 8 Since t h e s e n u m b e r s c a n n o t b e
(52-8)!
r e p e a t e d , t h e t o t a l n u m b e r of p a s s c o d e s u s i n g 3
8!
52!
11(0):
d i f f e r e n t e v e n d i g i t s is e q u a l t o
44!
8!
C'
01:
C ; -if
1
V
d
1 2 1
45
44!
n(0;
niA)
•L:'
=
(5^3)1
'.|
c. /
')(.
'V:
0
•
i r , :u
At.
45
fi(o;
/
2!
48
f-
.1 3
49
50
52
51
/
0
I
,,
/
46
niO)
45
niA)
2!
n(0)
n(^)=5-4-3
niA)
:
0
/
2.
•
(/
47
10.47
N o w d e l e n i u i i t ; the p i u D d b i l i t y ;
niA)
= 60
PiA)
niO)
T h e n u m b e r o f 3 digit p a s s c o d e s is 1 0 j o r 1000.
b e c a u s e t h e r e a r e 10 p o s s i b l e d i g i t s t o u s e i n 3
'P
PiA
s p a c e s , a n d r e p e a t i n g is a l l o w e d ,
'• i
M
If-13
PiA)
PiA
n(0)
PiA):
2
60
1000
/
0 "1 > ^ f
4,
1573
PiA)
2 315 502
3
PiA)
T h e probability that a hand consists of 4 hearts a n d 4
50
T h e p r o b a b i l i t y t h a t B r a y d o n ' s p a s s c o d e is m a d e
up
s p a d e s i s ^rzr^iz::
2315502
o r a b o u t 0,000 679,
3
of t h r e e different e v e n digits is —
, 0,06 o r 6 % .
10
5. P ( t r u c k o r y o - y c
df]
50
4. L e t A r e p r e s e n t a h a n d c o n t a i n i n g 4 hearts a n d 4
T h e r e a r e 13 h e a r t s a n d 13 s p a d e s i n a s t a n d a r d d e c k .
Therefore, the total n u m b e r of h a n d s that contain 4
h e a r t s a n d 4 s p a d e s i s 13C4 • 13C4, o r
n(A)
inCi)^.
=
L e t A r e p r e s e n t H a n s f l i p p i n g a c o i n o n h e a d s , a n d let
B r e p r e s e n t H a n s d r a w i n g a n 8,
13!
PiA)
9! • 4!
= \
2
PiA--.B)
4
1,'
niA,Of
-p
-1
lO Of 2
< J r
PiB)
52
I
'3
12
4
!1
P{AnB)
13
10
( 5 11 13)
niA)
5'' I f
=
'
P{B)
1
1
2
13
=
26
T h e probability that H a n s will flip a h e a d a n d d r a w a n
3~2
niA)^
PiAnB)
^
PiA)
1
1
PiB)
niA}-
'.) /
6. T h e s e e v e n t s a r e i n d e p e n d e n t .
( 1 3 - 4 ) ! - 4!
niA)
P(truck or yo-yo -
T h e probability that K a y l e e will w i n either a t o y truck
o r a y o - y o is 0,2. o r 2 0 % .
13!
= '
1
P(truck o r yo-yc
s p a d e s , a n d let O r e p r e s e n t all 8-card h a n d s .
8 is 4 , o r a b o u t 0 , 0 3 8 5 o r 3 , 8 5 % ,
26
13'
7. L e t P r e p r e s e n t m a k i n g a f r e e t h r o w ,
T h e t o t a l n u m b e r o f 8 - c a r d h a n d s i s 52C8.
a)
PiF)
= 0,7
P ( P n P ) = 0.7-0,7
P ( P r > F) = 0 , 4 9
T h e p r o b a b i l i t y J a r r o d will m a k e b o t h s h o t s is 0,49, o r
49%,
F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l u t i o n s
Manual
5-31
b|
PiF)
N o w determine the probability,
= 0.3
P{F'nF)
=
Q.3-0.3
P ( F ' n F) = 0 . 0 9
T h e p r o b a b i l i t y J a r r o d will m a k e b o t h s h o t s is 0 . 0 9 , o r 9 % .
c ) In t h i s c a s e , t h e r e a r e o n l y 3 p o s s i b i l i t i e s . T h e s e
are make both shots, m a k e one shot, and m a k e
neither shot.
Let O r e p r e s e n t J a r r o d m a k i n g e x a c t l y o n e s h o t ,
P ( 0 ) = 1 0 0 % - P ( P n P) - P ( P ' n F)
P(0)^
100%--49% - 9 %
P(0) = 42%
T h e p r o b a b i l i t y J a r r o d will m a k e o n e s h o t is 0 4 2 o r
42%.
d ) P - l 4 rf--pt(;sent J a r r o d m a k i n g a t l e a s t o n e s h o t ,
FIA)
U)ir'/.,~P(F'nF)
P(A)
P{H)
=
n{0)
15
276
T h e p r o b a b i l i t y t h a t G e o r g e d r a w s t w o h e a r t s is
5_
92 '
o r il>.iUl i; i)'A', Mt •) AA%.
9. 1 . I s- [f«|)i(-.(.i!l .i s e l a l a r m c l o c k , a n d let N
r e p r r s c n l ,ui a l a r m olc>ck t h a t is not set. Let L
ref l O ' . o r l M i q u f I boin(j l a t e .
P(h
/ ! - P(.S) 'Jf / \S)
P(I)
P(F
P- - 0 / 2 0 10
P(U
Pi.-;
L;
100:;.-9%
P>.A)
01%
p i o b o b i l i t y J a r r o d will m a k e at l e a s t o n e s h o t is
0.91 or 9 1 %
PiN
8 . Let H r e p r e s e n t t w o h e a r t s b e i n g d r a w n f r o m t h e
d e c k , a n d let O r e p r e s e n t all p o s s i b l e d r a w i n g s . T h e r e
a r e 6 h e a r t s in e v e r y e u c h r e d e c k . T h e r e f o r e , t h e
n u m b e r of w a y s to draw t w o hearts f r o m a Euchre
d e c k is aCi-
PIN.
u OP'
0 L) - f ' ! p )
P{S\L)
P{P • I i
PiL)
''.C.P)
0,072
.1 1 0 '^a n
PiSlL)
- U 10(",
0,268
PiS\L)
T h e p r o b a b i l i t y t h a t M i g u e l r e m e m b e r e d t o s e t his
a l a r m is a b o u t 0 , 2 6 9 o r 2 6 , 9 % ,
Chapter Review, page 367
(6-2)!
2!
6!
n{HP=
n(H) =
n{H)
Coin Flips
Winner
6^5y4!
HHHH
HHHT
Jie ^
4 ! 2!
n{H)
4! 2 1
HHTH
Tie
HHTT
Camila
2
mViH
Te
HTHT
Camila
H f 111
('amila
-15
d e c k is 24C2.
11(0):
.C,
(24
2 ) ! 2!
22! 2!
24 23 22!
n{0)
n{0)
n{0)
NffJ
Tie
fllhlll
Tie
'1IHHT
24!
24!
n{0)
Chloe
6 5
T h e n u m b e r of w a y s to d r a w 2 cards f r o m a E u c h r e
n{0)
1. a) O u t c o m e T a b l e
Camila
IHTII
Camite
THTl^"
Tie
flHH
Camila
tTHJ
Tie
TTfjd'
Tie
1 IT
Chloe
2 2 ! •2-1
24 23
P(Chloe wins)
2
16
2
276
P(Chloe wins)
P(Camila wins)
1
8
6_
16
P(Camila wins)
3
8
T h i s g a m e is not fair, e g , . C a m i l a h a s a b e t t e r c h a n c e
of winning.
5-32
C h a p t e r 5: P r o b a b i l i t y
b | It is e q u a l l y like t h a t a 1 o r a 2 will c o m e u p t h a n a
3 o r a 4 , T h e r e f o r e , t h e g a m e is fair, C o o p e r a n d
A l y s s a will h a v e a n e q u a l c h a n c e at w i n n i n g .
2 . T h e p r o b a b i l i t y is 0, b e c a u s e w h e n t h e t e n s i o n is at
1 0 0 % you c a n no longer t u m the pedal. At this point
t h e t e n s i o n is a t its m a x i m u m . T u r n i n g t h e t e n s i o n u p
to 1 3 0 % implies that tension c a n be increased b e y o n d
its m a x i m u m s e t t i n g . W h a t B o b ' s i n s t r u c t o r p r o b a b l y
m e a n s is t o a d j u s t t h e t e n s i o n s o t h a t it is 1.3 t i m e s
g r e a t e r t h a n it c u r r e n t l y is, w h i c h m a y b e p o s s i b l e .
3. a | P ( f e m a l e ) = 6 0 % ; P ( n o t f e m a l e ) = 4 0 %
T h e o d d s in f a v o u r o f t h i s p e r s o n b e i n g f e m a l e a r e
60 : 4 0 . or 3 ; 2
b) T h e o d d s a g a i n s t t h i s p e r s o n b e i n g f e m a l e a r e 2 : 3.
N o w determine the probability,
PiC)
niC)
niO)
T h e p r o b a b i l i t y C a m e r o n a n d W y a t t will b e c h o s e n f o r
p r e s i d e n t a n d s e c r e t a r y is ^ , o r a b o u t 0 . 0 2 7 8 o r 2 . 7 8 % .
36
9. Let M r e p r e s e n t a t e a m c o n t a i n i n g M a r i n a a n d
M a c K e n z i e , a n d let O r e p r e s e n t all p o s s i b l e t e a m s .
T h e n u m b e r of w a y s t h a t M a n n a a n d M a c K e n z i e c a n
b e p l a c e d o n t h e t e a m is 4P2, a n d t h e n u m b e r o f w a y s
t o fill t h e o t h e r 2 s p o t s o n t h e t e a m is 9P2. T h e r e f o r e ,
4. T h e o d d s a g a i n s t rain t o m o r r o w a r e
the total n u m b e r of t e a m s containing Marina a n d
( 1 0 0 % - 7 0 % ) ; 7 0 % . T h i s is e q u a l t o 3 0 : 7 0 . o r 3 : 7.
5. a ) If Keir fell t w i c e , t h e n h e d i d n ' t fall t h e o t h e r f i v e
t i m e s . T h e r e f o r e , t h e o d d s in f a v o u r of h i m f a l l i n g a r e
2 : 5.
M a c K e n z i e is 4P2 • 9P2.
P • p
niM):
4 2 9' 2
9!
4!
niM)
(4 - 2)1 (9-^2)1
b) T h e o d d s a g a i n s t Keir f a i l i n g a r e 5 : 2 .
6. N o , s h e IS n o t c o r r e c t , e t
c u l a t e t h e o d d s , it
s h o u l d b e o d d s a g a i n s t : o d d s for. If A n a n a h a s
scored 6 goals on 30 shots, then she did not score
24 times. Therefore, the odds against her s c o n n g a
g o a l a r e 2 4 ; 6, or 4 : 1, not 4 : 5.
7. Let P r e p r e s e n t A v e n l l g e t t i n g a n A in t h e first
c o u r s e , a n d let S r e p r e s e n t A v e r i l l g e t t i n g a n A in t h e
s e c o n d conrqA
P(T
PiS)
p(
Pi^
%
niM)
7!
4 3 2 ! 9 8 7!
2!
7!
f7(M)-4-3-9-8
n{M)-
864
T h e total n u m b e r of p o s s i b l e t e a m s is 11P4.
6-11
P(S) :
2,'/72
^-M
.(M) =
'
2!
^
P(S) = 35,294,..%
niO) =
^
^ '
(11-4)!
niO)
Avenll should take the second course, because the
p r o b a b i l i t y of g e t t i n g a n A is h i g h e r .
n(0)
11!
7!
I 1 l u '. \
"
8. Let C r e p r e s e n t C a m e r o n a n d W y a t t b e i n g c h o s e n
f o r p r e s i d e n t a n d s e c r e t a r y , a n d let O r e p r e s e n t all
fi(O)
p o s s i b l e c h o i c e s . B e c a u s e o r d e r is i m p o r t a n t , t h e r e
n(0) = 7 9 2 0
are 2 w a y s that C a m e r o n and Wyatt can be chosen
N o w determine the probability,
f o r p r e s i d e n t a n d s e c r e t a r y . T h e total n u m b e r o f w a y s
t h a t 2 p e o p l e c a n b e c h o s e n is 9P2.
PiM)
niO)=A^_
niM)
n{0)
864
PiM)
niO)
(92)1
7920
r7(0)
niO)
9-8-7!
7!
n{0)
9 8
niO)
72
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
T h e p r o b a b i l i t y M a n n a a n d M a c K e n z i e will b e c h o s e n t o
b e o n t h e r e l a y t e a m is 3 - , o r a b o u t 0 , 1 0 9 o r 1 0 , 9 % ,
55
1 0 , L e t A r e p r e s e n t a p a s s w o r d c o n t a i n i n g t h e letters
A . R, a n d T, a n d let O r e p r e s e n t all p o s s i b l e
passwords.
5-33
a) T h e r e a r e 3! w a y s o f a r r a n g i n g t h e l e t t e r s A , R, a n d
T , a n d t h e r e a r e 10P4 w a y s of a r r a n g i n g t h e f o u r d i g i t s .
N o w determine the probability.
P{A)
=
P{A)
=
Therefore, the n u m b e r of passwords that contain the
niA)
n{0)
l e t t e r s A , R, a n d T is 3! • 10P4.
n{A) =
60000
3\;,P,
175760000
10!
n ( A ) = 3!
n{A) =
3
PiA)
(10-4)1
8788
T h e p r o b a b i l i t y t h a t a p a s s w o r d c h o s e n at r a n d o m will
10-9-8
3-2A
c o n t a i n t h e l e t t e r s A , R, a n d T is
7-6!
n(/\)-3-2-10-9-8-7
T h e total n u m b e r o f p a s s w o r d s c o n t a i n i n g t h r e e
c a p i t a l letters f o l l o w e d b y f o u r d i g i t s , w i t h o u t
r e p e t i t i o n , is aePa • 10P4n(0)=,,P3-,,P,
26!
10!
'
(26-3)!
(10-4)!
0.000 341 or 0 . 0 3 4 1 % .
1 1 . a ) T h e s e e v e n t s a r e not m u t u a l l y e x c l u s i v e ; e . g . ,
There are prime numbers that are also odd numbers.
b) T h e s e e v e n t s a r e m u t u a l l y e x c l u s i v e ; e . g . , Y o u
c a n n o t roll a 6 a n d a n 8 a t t h e s a m e t i m e .
c) T h e s e events are mutually exclusive; e.g., You
cannot eat a p e a c h a n d an apple at the s a m e time.
a spade.
26! 10!
^ - ^
26 25 2 4 - 2 3 !
n{0)
or about
1 2 . a ) L e t A r e p r e s e n t a f a c e c a r d , a n d let B r e p r e s e n t
.
n(0) =
3
8788
=
10 9 8 - 7
23!
6!
6!
n ( 0 ) = : 2 6 25 24 10-9 8 7
N o w determine the probability.
P{A)
n{A)
n(0)
3 2 10 9 - 8 - 7
PiA)
2 6 2 5 2 4 10 9 8 - 7
b) T h e s e e v e n t s a r e n o t m u t u a l l y e x c l u s i v e . T h i s is
b e c a u s e there are 6 cards that are both face cards
a n d s p a d e s (2 j a c k s , 2 q u e e n s a n d 2 k i n g s ) .
3-2
PiA)
26-25
24
1
PiA)
24
26-25 4
0)
P(B)
P(A).-
48
1
PiA)
2600
T h e p r o b a b i l i t y t h a t a p a s s w o r d c h o s e n a t r a n d o m will
48
PiA uB)
1
c o n t a i n t h e letters A , R, a n d T is
P{AnB)^
= PiA) + PiB) - PiA n B )
, or a b o u t
24
2600
0.000 3 8 5 or 0 . 0 3 8 5 % .
b) T h e r e a r e 3! w a y s o f a r r a n g i n g t h e letters A , R, a n d
T, a n d t h e r e a r e lO'^ w a y s o f a r r a n g i n g t h e f o u r d i g i t s .
Therefore, the n u m b e r of p a s s w o r d s that contain the
letters A , R, a n d T is 3!
loJ
n{A) = 3\ - IO''
n{A) = 3-2
1 - 10 0 0 0
12
48
PiAuB)
=
30
48
T h e p r o b a b i l i t y t h a t H u n t e r will d r a w a f a c e c a r d or a
30
s p a d e is — , 0 . 6 2 5 o r 6 2 . 5 % .
48
n{A) = 6 0 0 0 0
T h e total n u m b e r o f p a s s w o r d s c o n t a i n i n g t h r e e
c a p i t a l letters f o l l o w e d b y f o u r d i g i t s , a l l o w i n g
r e p e t i t i o n , is 2 6 ^ - 10"*
n ( 0 ) = 2 6 ^ - IO''
n{0)=
5-34
175 760 000
C h a p t e r 5: P r o b a b i l i t y
1 3 . a ) Let G r e p r e s e n t e x e r c i s i n g o n S u n d a y , a n d let
S represent shopping on Sunday.
15. e . g . . O u t o f 5 0 retail o u t l e t s , 19 a r e h o l d i n g s a l e s
this m o n t h . 15 o u t l e t s sell o n l y s u s t a i n a b l y
manufactured items, and 6 of these are holding sales
this m o n t h . W h a t is t h e p r o b a b i l i t y t h a t a retail o u t l e t
sells s o m e non-sustainably m a n u f a c t u r e d items a n d
is h a v i n g a s a l e ?
M = {outlets selling sustainably manufactured items}
S = {outlets having sales}
b) T h e t w o e v e n t s a r e not m u t u a l l y e x c l u s i v e ,
b e c a u s e t h e p r o b a b i l i t y t h a t M y a will b o t h e x e r c i s e
a n d s h o p o n S u n d a y is not e q u a l t o 0 (it is e q u a l t o
0.2).
c ) P ( G ) = 0.5
P ( S ) = 0.3
P ( G n S) = 0.2
P ( G u S) = P ( G ) + P ( S ) - P ( G n
S)
P ( G u S ) = 0.5+ 0.3-0.2
22
S o l u t i o n : Let R r e p r e s e n t retail o u t l e t s h a v i n g s a l e s
t h a t sell s o m e n o n - s u s t a i n a b l y m a n u f a c t u r e d i t e m s .
F r o m t h e V e n n d i a g r a m w e s e e 13 retail o u t l e t s t h a t
sell s o m e n o n - s u s t a i n a b l y m a n u f a c t u r e d i t e m s a r e
having sales.
P ( G u S ) = 0.6
T h e p r o b a b i l i t y t h a t M y a will d o o n e o f t h e s e a c t i v i t i e s
o n S u n d a y is 0.6, or 6 0 % .
14. e . g . . S u p p o s e 6 s t u d e n t s c a n b o t h ski a n d s w i m .
W h a t is t h e p r o b a b i l i t y t h a t a r a n d o m l y s e l e c t e d
s t u d e n t c a n n o t ski o r s w i m . Let W = { s t u d e n t s w h o
s w i m } a n d K = { s t u d e n t s w h o ski}.
PiR)
= ~
T h e p r o b a b i l i t y is —
50
50
16. Let B r e p r e s e n t a b l a c k s o c k , a n d let H / r e p r e s e n t
a white sock. Let P represent Parker pulling a black
s o c k t h e n a w h i t e s o c k , a n d let G r e p r e s e n t P a r k e r
pulling a white sock then a black sock. There are two
w a y s to d r a w a w h i t e s o c k a n d a b l a c k s o c k .
PiB)
18
4
P{B)-
P{W\B)-
P{W)
10
5
8
40
P{F)-
" 153
P ( G ) = PiW)P{B\W)
P(G) =
5
8
917
P(G) =
40
153
17
24 ~ 8
10
" 9 1 7
9
PiBlW)-
P{B)P{W\B)
4
P(P).
17
18
P{W)
P ( P ) .=
9
10
S o l u t i o n ; T h e total n u m b e r o f s t u d e n t s w h o c a n s k i ,
s w i m o r ski a n d s w i m is 13 + 6 + 2 = 2 1 . T h e r e a r e
3 students w h o cannot ski or s w i m . T h e probability
t h a t a r a n d o m l y s e l e c t e d s t u d e n t c a n n o t s k i o r s w i m is
0.26 or 2 6 % .
0 . 1 2 5 or 1 2 . 5 % .
P{BnW)
= P{F) + P(G)
D/D
M/\
40
40
P(B nW) = — +
153
153
80
P{BnW)
153
T h e p r o b a b i l i t y t h a t P a r k e r will pull o u t a pair of
m i s m a t c h e d s o c k s is
, or about 0.523 or 5 2 . 3 %
153
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5-35
1 7 . Let L r e p r e s e n t a p l a n e l e a v i n g f r o m W i n n i p e g o n
2 1 . Let F r e p r e s e n t p a s s i n g F r e n c h , a n d let C
t i m e , a n d let A r e p r e s e n t a p l a n e a r r i v i n g in C a l g a r y
represent passing chemistry,
on time.
a)
P(L) = 0 70
PiL n
m \ L )
P{FnC)
0.56
A)
P ( P ) = 0,7
PiC) = 0 6
_
P{LnA)
:
^^^^
=
PiF)-PiC)
P ( P n C ) = 0,7-0,6
P(P n C) = 0.42
T h e p r o b a b i l i t y T a n y a will p a s s b o t h F r e n c h a n d
PiA 1 L)
0.56
c h e m i s t r y is 0 , 4 2 , o r 4 2 %
0.70
b)
0.8
P(A I L)
PiC)
PiFnC)
T h e p r o b a b i l i t y t h a t a p l a n e will a r r i v e in C a l g a r y o n
t i m e , g i v e n it left W i n n i p e g o n t i m e , is 0.8, o r 8 0 % .
P{Fn
= 0,4
=
PiF)P{C')
CO = 0,7
0,4
P ( P n CO = 0 , 2 8
T h e p r o b a b i l i t y T a n y a will p a s s F r e n c h b u t fail
18. Let R r e p r e s e n t t h e first c a r d d r a w n b e i n g r e d ,
a n d let B r e p r e s e n t t h e s e c o n d c a r d d r a w n b e i n g
black.
PiR)
=
20
P{RnB)
40
1
20
2
2
39
20
20
1
P{R)
P{RnB)
PiRnB)
p(b|r) =
78
39
T h e p r o b a b i l i t y t h o first c a r d is r e d a n d t h e s e c o n d
20
c a r d IS b l a c k is — , o r a b o u t 0 . 2 5 6 o r 2 5 . 6 % .
78
19. Let S r e p r e s e n t u s i n g a stair m a c h i n e , a n d let B
represent going to a body sculpt class.
PiS) =
P{S
l
P{B)
- B)
P{S)
\
P{B)
c h e m i s t r y is 0 , 2 8 , o r 2 8 % ,
c)
fif ) - 0 3
P(l
, . C1
Pp
'. , f , )
P{F)
PiC)
0 3 • 0,4
Pp
• F- } " f i 12
T h e p t o b a b i l i l y T a n y a will fail b o t h F r e n c h a n d
c h e m i s t r y IS 0 , 1 2 , or 1 2 % ,
C h a p t e r T a s k , page 369
A I chose Gin Rummy,
B . t or o n e r o u n d , I h a v e A v , 2 v . 3 v , m, 6#, 8 * , 9 # ,
I Of, J # , a n d m in m y h a n d . I k n o w t h a t m y o p p o n e n t
has picked up the K # a n d the 7#. I also k n o w that he
h a s d i s c a r d e d t h e 9# a n d t h e J v , T h e M is c u r r e n t l y
f a c e u p . T h e r e a r e still 2 0 u n t u r n e d c a r d s in t h e d e c k ,
C . T h e r e a r e . at m o s t , s e v e n h e a r t s left in t h e d e c k ,
i n c l u d i n g t h e u n k n o w n c a r d s in m y o p p o n e n t ' s h a n d , I
could pick up the J v , hoping to get either another jack
or the
d u r i n g a later p l a y , o r I c o u l d d r a w a f a c e d o w n card. T h e conditional probability of drawing the
2
1
or
, This would
20
10
strengthen my hearts, potentially getting a longer run;
b u t if I d r e w t h e 5 v a n d d i d n o t g e t t h e 4 ¥ d u n n g a
later t u r n , m y h a n d w o u l d n o t b e s t r e n g t h e n e d . T h e
p r o b a b i l i t y of d r a w i n g o n e of t h e s e c a r d s n o w a n d o n e
4 ¥ o r 5 ¥ is, at m o s t .
PiSr^B)
^
T h e p r o b a b i l i t y t h a t P e y t o n will u s e a s t a i r m a c h i n e
a n d t a k e a b o d y s c u l p t i n g c l a s s w h e n s h e n e x t visits
t h e g y m is
. or a b o u t 0 . 1 6 7 o r 1 6 , 7 % ,
6
on the next turn w o u l d be. at most,
f 1 If , 1- 1or
y l O ^ y Iyy
2 0 . If t h e e v e n t s a r e i n d e p e n d e n t , t h e n t h e p r o d u c t of
t h e i n d i v i d u a l p r o b a b i l i t i e s will e g u a l t h e p r o b a b i l i t y o f
both the events occurnng.
P{A nB)
= P{A)-
PiB)
0,3 = 0,5 • 0,6
0,3 = 0.3
Yes. these events are independent.
-(I u s e d 19 b e c a u s e I a s s u m e d t h a t m y o p p o n e n t
190
w o u l d p i c k u p m y d i s c a r d e d c a r d , I w o u l d u s e 18 if h e
d r e w a c a r d f r o m t h e d e c k , ) If I pick u p t h e J ¥ , t h e n
there are possibly three cards that would strengthen
m y h a n d r e m a i n i n g in t h e d e c k : t h e Q v , t h e J * , a n d
t h e J#. T h e p r o b a b i l i t y t h a t I will g e t o n e of t h o s e
3
It m a k e s
c a r d s in m y n e x t d r a w is, a t m o s t ,
s e n s e t o pick u p t h e Jw. I still n e e d t o d i s c a r d o n e of
t h e c a r d s in m y h a n d . S i n c e m y o p p o n e n t t h r e w a w a y
the 9 * , I a m going to g u e s s that he m a y not have
c a r d s n e a r t h e 9 f t o m a k e a r u n . s o it m a k e s s e n s e t o
d i s c a r d t h e 10#.
5-36
C h a p t e r 5: P r o b a b i l i t y
D. O n e s t r a t e g y is t o o b s e r v e w h i c h c a r d y o u r
o p p o n e n t collects and to think about w h y he or s h e
w a n t s t h i s c a r d . D o e s y o u r o p p o n e n t w a n t it t o m a k e
a set or a r u n ? If s o , y o u m i g h t w a n t t o p r e v e n t y o u r
o p p o n e n t f r o m getting cards that w o u l d allow him or
h e r t o c o m p l e t e t h e set. A n o t h e r s t r a t e g y is t o c o l l e c t
a c a r d y o u d o n o t really n e e d t o f o o l y o u r o p p o n e n t
into t h i n k i n g t h a t y o u a r e l o o k i n g for t h i s t y p e o f c a r d .
3-5 Cumulative Review, page
c ) 2 7 - 14 = 13, s o 13 s t u d e n t s like o n l y a p p l e s .
2 7 - 2 6 = 1, s o 1 s t u d e n t likes o n l y o r a n g e s .
2 7 - 1 3 - 1 = 1 3 , s o 13 s t u d e n t s like o r a n g e s a n d
a p p l e s . 14 s t u d e n t s like o n l y a p p l e s o r o n l y o r a n g e s .
4. a) e . g . , o d d w h o l e n u m b e r s l e s s t h a n 1 0 0 a n d e v e n
w h o l e n u m b e r s less than 100
b) e . g . , o d d w h o l e n u m b e r s l e s s t h a n 1 0 0 a n d p n m e
n u m b e r s less than 100
373
1. a) e . g . , M a n i t o b a , Q u e b e c , P E I , N e w B r u n s w i c k ,
Nova Scotia
b) e . g . , X is a n integer, s o x = { . . . , - 2 , - 1 , 0, 1 , 2 , 3, 4 ,
5, . . . } . fa is 10 t i m e s m o r e t h a n x .
5. Let S = { c a m p e r s w h o l e a r n e d s i n g i n g } ,
D = {campers w h o learned dancing},
A = {students w h o learned acting}
10 • 1 = 1 0 , 10
n{S n D ) = 2 1 , n ( D nA)
2 = 2 0 , 10 • 3 = 3 0 , 10
4 = 40,
10 • 5 = 5 0 . S o , 1 0 , 2 0 , 3 0 , 4 0 , 5 0 a r e f i v e e l e m e n t s .
2 . a)
T h e n : n ( S ) = 3 5 , n{D) = 3 8 , n{A) = 3 3
n{Dij
SuA)
= 2 3 , r7(S n
A)
= 56
By the Pnnciple of Exclusion and Inclusion:
n{Du
Su
A)
= n { S ) + n ( D ) + n{A) - n{S nD)+
n{D nA)~
n{S n
A)
n{DnSnA)
5 6 = 3 5 + 3 8 + 3 3 - 2 1 - 2 3 - 18 + r7(D
S r^ / \ )
12 = n ( D n S n / \ )
! i.'d-.iy
Si,in:f-jy
b) A a n d B h a v e n o c o m m o n e l e m e n t s , s o t h e y a r e
disjoint sets.
c ) i) f a l s e ; e . g . , A a n d B a r e d i s j o i n t s e t s , s o A is not a
subset of 6 .
ii) t r u e ; C is e n t i r e l y i n s i d e A, s o it a s u b s e t oi A.
iii) t r u e ; B c o n t a i n s all t h e e l e m e n t s t h a t a r e n o t in A,
s o B is t h e i n v e r s e o f / A .
iv) f a l s e ;
n{A) + n{B) + r7(C) = 5 + 2 + 2
n{A) + r?(B) + n{C) = 9
b u t n{U) = 7 # 9.
T h i s is b e c a u s e A i n c l u d e s C, s o r7(C) is c o u n t e d
twice.
S o , 12 c a m p e r s l e a r n e d s i n g i n g , d a n c i n g , a n d a c t i n g .
6. a) Y e s , it is t r u e ; a n e g a t i v e n u m b e r m u s t b e l e s s
than zero.
b) If a n u m b e r is l e s s t h a n z e r o , t h e n it is n e g a t i v e .
Y e s . e . g . . All n u m b e r s l e s s t h a n z e r o m u s t b e
negative.
c ) If a n u m b e r is not n e g a t i v e , t h e n it is not l e s s t h a n
z e r o . Y e s . e . g . . All n u m b e r s t h a t a r e n o t n e g a t i v e a r e
e i t h e r z e r o o r p o s i t i v e , a n d all o f t h e s e n u m b e r s a r e
not l e s s t h a n z e r o .
d) If a n u m b e r is n o t l e s s t h a n z e r o , t h e n it is not
n e g a t i v e . Y e s . e . g . . All n u m b e r s n o t l e s s t h a n z e r o
a r e e i t h e r z e r o o r p o s i t i v e , a n d t h e y a r e all n e g a t i v e .
e) Y e s . e . g . . B o t h t h e s t a t e m e n t itself a n d a n d t h e
converse are true so the statement can be wntten as
a biconditional.
3. a)
b) T h e r e a r e 3 5 s t u d e n t s in a l l , a n d 8 like n e i t h e r fruit,
s o 3 5 - 8 = 2 7 s t u d e n t s like o r a n g e s o r a p p l e s .
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5-37
10!
7. a )
c)
Coin
Qm
^10-9-8-7y6!
6131
10!
6!3!
^10-9-8-7
6 ! 3 ! ^ " ~~3-2-1
10!
840
6!3!
1.
d)
.,R
(0
t.).;l%
!.)!
Ot
,/2
^
4'
J
F
/
t
0
1'
4'
P, - 9 B / 6 5
heads
P
tails
i n 120
9!
(9-5)!5!
9!
b) R o l l i n g a d i e h a s 6 o u t c o m e s : 1 , 2, 3, 4 , 5. 6;
'
t o s s i n g a c o i n ti-:c 2 i u d c o m e s : h e a d s , tails
By the F u n d a m e n t a l C o u n t i n g Pnnciple, there are
6
2 = 12 c o m b i n e d o u t c o m e s , w h i c h is h o w m a n y
=
4!5!
0 8 7 0 5!
(2
415!
t h e r e a r e in t h e t r e e d i a g r a m .
0 S-7-6-
8. a ) A l i c e n c e p l a t e w o u l d b e o f t h e f o r m LLL
NN,
w h e r e L r e p r e s e n t s a letter, a n d N r e p r e s e n t s a
n u m b e r . A n y o n e o f t h e 2 6 l e t t e r s c o u l d b e u s e d in a
4 2 2
C,
9 2 7
,,C.
126
s p o t , a n d s o c o u l d a n y o n e of t h e 10 n u m b e r s .
T o t a l l i c e n c e p l a t e s = 26
26
26
10
10
Total licence plates = 1 757 600
12!
'Mil
b) T o t a l l i c e n c e p l a t e s w i t h o u t r e p e t i t i o n
= 2 6 • 2 5 - 2 4 - 10 ' 9
= 1 404 000
Total licence plates with repetition
il2
8)!8!
12"
> 8 J
I'd!
M2
12 11 10 9 3 !
, 3
418!
= total licence plates - total licence plates without
' ^2 1 ^ 12 n j i O
repetition
= 1 757 600 - 1 404 000
= 353 600
, 8;
12 ;
Therefore, 353 600 different plates use the s a m e
letter o r s a m e n u m b e r m o r e t h a n o n c e .
I
9
4 3 2 " "
115
9
8;
f12^
9. a ) 1 2 ! = 4 7 9 0 0 1 6 0 0
495
8!
P.
(8-8)1
10. a ) T h e r e a r e 1 5 ! w a y s , o r 1 3 0 7 6 7 4 3 6 8 0 0 0
8!
ways.
0!
b) O u t o f 15 m e m b e r s , 3 c a n b e c h o s e n , w i t h r e g a r d
8!
f o r o r d e r , in 15P3 = 2 7 3 0 w a y s .
c ) O u t of 15 m e m b e r s , Jill, a n d 2 o t h e r m e m b e r s c a n
R = 40 320
b e c h o s e n in 14C2 = 9 1 w a y s .
11. T h e r e w e r e 13 + 5 = 18 g a m e s p l a y e d , s o t h e r e
a r e 18C13 = 8 5 6 8 w a y s in w h i c h 13 w i n s c o u l d h a v e
occurred.
5-38
C h a p t e r 5: P r o b a b i l i t y
1 2 . a ) T h e r e a r e 7 s q u a r e s to t h e e a s t , a n d 3 s q u a r e s
s o u t h , 7 + 3 = 10. s o t h e r e a r e
120
c ) T h e r e will b e m o r e p a r e n t s t h a n s t u d e n t s if t h e r e
are
• 6 parents and no students:
7C6 = 7 w a y s
different routes.
• 5 parents and 1 student:
b) In t h e first g r i d , t h e r e a r e 3 s q u a r e s t o t h e e a s t , a n d
2 s o u t h . In t h e o n e s q u a r e , t h e r e is o n e s q u a r e t o t h e
e a s t a n d o n e t o t h e s o u t h . In t h e b o t t o m g r i d , t h e r e
are 3 s q u a r e s to the east a n d 3 to the south. So. there
are |
|-2
: 4 0 0 different routes.
3
13
n(n^^.l)(n....2)(r>...-3)(n
4)!
(n....4)!4!
^ri--n)(n--2)in3)1
(n
3)\
2 ^ " _ 2 2 i 2 „ , „ . , ) , , , , )
4!
•
4"l)(n--2)(n
3) = 24n(n
n
l){n2)
3 - 24
n = 27
14. a) T h e r e a r e 7 + 9 = 16 p e o p l e in t o t a l .
If 3 a r e p a r e n t s a n d 3 a r e s t u d e n t s , t h e n :
7C3 • 9C3 = 35 - 8 4 o r 2 9 4 0 w a y s in t h i s c a n h a p p e n ,
b) If t h e r e is at l e a s t 1 s t u d e n t , t h e n t h e r e c a n b e :
• 5 parents and 1 student:
7 C 5 • 9C1 = 2 1 - 9 o r 1 8 9 w a y s
• 4 parents and 2 students:
7C4 • 9C2 = 35
36 or 1260 w a y s
7 C 5 - 9 C 1 = 21 - 9 or 189 w a y s
• 4 parents and 2 students;
7C4 - 9C2 = 35 3 6 o r 1 2 6 0 w a y s
Total n u m b e r of w a y s
= 7 + 189 + 1260
= 1456
T h e r e a r e 1 4 5 6 w a y s in w h i c h t h e c o m m i t t e e c a n
have more parents than students,
1 5 . T h e r e a r e 4 a c e s in a d e c k , s o t h e n u m b e r o f
w a y s in w h i c h a n a c e c a n b e in t h e h a n d is 4C1 = 4 ,
T h e r e a r e 4 t e n s in t h e d e c k , s o t h e n u m b e r o f w a y s
in w h i c h a n a c e c a n b e in t h e h a n d is 4C1 = 4 ,
T h e r e a r e 5 0 o t h e r c a r d s in t h e d e c k , a n d t h e w a y s in
o n e of t h e m c a n b e in t h e h a n d is 50C1 = 5 0 ,
Multiply these figures,
4 • 4
50 = 8 0 0
T h e r e a r e 8 0 0 w a y s in w h i c h a t h r e e - c a r d h a n d c a n
h a v e (at l e a s t ) o n e a c e a n d (at l e a s t ) o n e t e n ,
16. I m a d e t a b l e of t h e p o s s i b l e r e s u l t s , I p u t a n "A" in
t h e spf--*- v / b r r r A m b c r ¥/cu!d w i n 1 " J " in th.- s p o t s
v/lK-ro« Jef ki-iyr, viouid win ri V' c> 2 in tht-' s p o t s
vjiv-tv -leil'-"- • wonlfl w i u
I)!.'
1
Liir
1
2
3
4
J
• 3 parents and 3 students:
7 C 3 • 9 C 3 = 35
84 or 2 9 4 0 w a y s
• 2 parents and 4 students;
7C2 • 9C4 = 21 • 126 o r 2 6 4 6 w a y s
• 1 parent and 5 students;
7C1 - 9C5 = 7 • 1 2 6 o r 8 8 2 w a y s
• 0 parents and 6 students;
9C6 = 8 4 w a y s
Total n u m b e r of w a y s
= 189 + 1260 + 2940 + 2646 + 882 + 84
= 8001
T h e r e a r e 8 0 0 1 w a y s in w h i c h t h e c o m m i t t e e c a n
h a v e at l e a s t 1 s t u d e n t .
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
A m b e r would win 9 out 36 times, Jackson would win 9
o u t of 3 6 t i m e s . N e i t h e r will w i n 18 o u t of 3 6 t i m e s .
T h e g a m e is fair, b e c a u s e b o t h p l a y e r s h a v e a n e q u a l
chance of winning.
17. a ) T o d e t e r m i n e t h e o d d s in f a v o u r w h e n t h e o d d s
a g a i n s t a r e k n o w n , s w i t c h t h e t e r m s in t h e ratio. T h e
o d d s a g a i n s t a r e 1 : 2 5 . s o t h e o d d s in f a v o u r a r e
25 ; 1
b) T o d e t e r m i n e t h e p r o b a b i l i t y in f a v o u r of a n e v e n t
w h e n t h e o d d s in f a v o u r a r e k n o w n , a d d t h e t e r m s , a n d
u s e t h e s u m f o r t h e ratio. T h e o d d s in f a v o u r a r e 2 5 : 1 .
s o t h e p r o b a b i l i t y is
or a b o u t 0 , 9 6 ,
1 + 25
26
18. S i n c e e a c h s t u d e n t is e g u a l l y likely t o w i n . t h e n
S o n y a is j u s t a s likely t o f i n i s h in t h e t o p t h r e e (the t o p
half) a s in t h e b o t t o m t h r e e (the b o t t o m h a l f ) . T h e
p r o b a b i l i t y S o n y a will f i n i s h in t h e t o p t h r e e is 0,5.
5-39
1 i . -3' I l-.-re a r e 52C13 = 6 3 5 0 1 3 5 5 9 6 0 0 w a y s in w h i c h
22 a ) Jon • - M ' W M
13 c a r d s c a n b e s e l e c t e d f r o m a d e c k o f 5 2 c a r d s
• .sht; W i n s hr.th f l a m e s ;
There
is o n l y o n e w a y in w h i c h all o f t h e s e 1 < ards c a n b e
it l e a s t o n e g a m e in t h r e e w a y s ;
/ •( Win br.th ( j a n u rd
h e a r t s T h e p r o b a b i i i t y t h a t a p l a y e r is d e a l t all h e a r t s is
- / ' { w i n g a m . : I) / ( w i n g a m e 2 | w o n g a m e 1)
1
635 0 « 559 600 '
^ other players's cards
,39
" "ail p o s s i b l e t^ands"
52
13^
= AL or 0 , 3 1 2 5
16
1 677 106 6 4 0
• She wins g a m e 1 and loses g a m e 2;
635 013 559 600
T h e p r o b a b i l i t y is
1
P. '
(48
all k i n g s
b)
1,1
/ ! J
1 6 7 7 106 6 4 0
P ( w i n g a m e 1 , lose g a m e 2 )
= P ( w i n g a m e 1) P ( l o s e g a m e 2 | w o n g a m e I )
635 013 559 600
2 0 . a) e g . , c h o o s i n g a n a p p l e o r a p e a r f r o m a b o w l
2 (
of fruit
b) e g , , c h o o s i n g 2 b l u e m a r b l e s f r o m a b a g
containing 7 blue and 3 red marbles, without
replacement
8
I )
(3
2 I'l
c ) e g , , rolling a s t a n d a r d die a n d g e t t i n g 4 , t o s s i n g a
8
or 0 , 1 8 7 5
coin and getting heads
16
2 1 . I m a d e a t a b l e of t h e s u m s . T h e r e a r e 3 6 p o s s i b l e
• S h e lose g a m e 1 and wins g a m e 2;
results
Ihv. I '
3
4 '
1
2
3
4
5 ^
f)
5
6
^5
6
H
7
7
H"
P ( l o s e qoti»e 1 w i n g a m e 2 )
P I lose qam(
1
H
5
6
•P
9
i"o
' 9
10
1 1
1) p ( w i n g a m e 2 | lost g a m e I )
'I
9
10
1 1
2 1 ~Z
4
f I
1
1
f 1
rl
a) T h e s u m is 3 or 12 o n t h r e e o c c a s i o n s .
Of 0 , 1 2 5
T h e p r o b a b i l i t y of this is
or about 0,0833,
36
b) T h e s u m is o d d o r t h e s u m is g r e a t e r t h a n 7 o n
27
2 7 o c c a s i o n s . T h e p r o b a b i l i t y of this is ~ - or 0 , 7 5 ,
36
c ) T h e r e a r e 6 o c c a s i o n s o n w h i c h t h e first d i e is 1, O f
t h e s e , t h e r e a r e t h r e e o c c a s i o n s o n w h i c h t h e s u m is
o d d . T h e p r o b a b i l i t y of t h i s is | o r 0,5,
6
d) T h e r e a r e 2 o c c a s i o n s o u t of 3 6 w h e n t h e first d i e is
l e s s t h a n 3 a n d t h e s e c o n d is g r e a t e r t h a n 5, T h e
p r o b a b i l i t y of t h i s is
or a b o u t 0 , 0 5 5 6 ,
T h e t o t a l of t h e s e p r o b a b i l i t i e s is
0,3125 + 0,1875 + 0.125 = 0,625
T h e p r o b a b i l i t y J a n will w i n a t l e a s t o n e g a m e is
0,625,
b)
P ( Stan wins both g a m e s )
= 1 - P I J a n w i n s at least o n e g a m e )
1-0,625
- 0,375
23. a) There are 24 cards, of w h i c h 4 are q u e e n s . T h e
p r o b a b i l i t y of d r a w i n g a g u e e n o n t h e first d r a w is
36
--24
If t h e c a r d is r e p l a c e d , t h e p r o b a b i l i t y of
6
d r a w i n g a q u e e n o n t h e s e c o n d d r a w is t h e s a m e ,
So, P(2 queens) = ^ I
6 6
5-40
^,
6
or a b o u t 0 , 0 2 7 8 ,
36
C h a p t e r 5: P r o b a b i l i t y
b ) T h e p r o b a b i l i t y of d r a w i n g a q u e e n o n t h e first d r a w
r e m a i n s t h e s a m e , ^ . If this c a r d is not r e p l a c e d ,
6
t h e n t h e r e is o n e l e s s q u e e n t o d r a w a n d o n e l e s s
c a r d t o d r a w f r o m . T h e p r o b a b i l i t y of d r a w i n g a g u e e n
3
= — . So,
24 • 1
23
12
3
or about 0.0217.
23
552
3'
/ c
4-1
o n t h e s e c o n d d r a w is
P(2 queens) =
4
24
Fc,'!.' '
('^- ..',n fiu n e n d a n t in 15 120 w a y s .
"
Chapter 5 Diayrujstic Test, 1 R page 343
1 i i I / / t . c th.- - ,..«.-orsal s e t .
^
24
( 2 4 - 5 ) ! - 5!
2t
,'3
r
5 ^
O
C 1 3
i n '
/:
J
20
2
2.1'
7;
V
^
T h e s t u d e n t s c a n b e s e l e c t e d in 4 2 5 0 4 w a y s ,
7. A g r e e , e . g . . T h e o u t c o m e of t o s s i n g t h e c o i n h a s
n o b e a r i n g o n t h e o u t c o m e of t o s s i n g t h e d i e . T h i s
means these are independent events.
7
70
2. a) T h e n u m b e r o f w a y s t o roll a 5 is 1; t h e n u m b e r
of o u t c o m e s for t h e die is 6. T h e n u m b e r o f w a y s t o
(• s s I'l.; r-j 1; t h e n u m b e r of o u t c o m e s is 2.
Review of Terms and Connections,
T R page 345
1 , a) ix) If o n e t h i n g c a n b e d o n e in f i v e w a y s a n d
a n o t h e r t h i n g c a n b e d o n e in t h r e e w a y s , t h e n b o t h
t h i n g s c a n b e d o n e in 5 3 o r 15 w a y s , a c c o r d i n g t o
the Fundamental Counting Principle
b) ii) T w o s e t s t h a t h a v e n o e l e m e n t s in c o m m o n a r e
mutually e x c l u s i v e ,
Jl
12
b} ' he . . n i n b e r o f w a y s t o roll a n e v e n n u m b e r is 3;
t h e n u m b e r o f o u t c o m e s is 6 T h e n u m b e r o f w a y s t o
tos<5 a h e a d P 1 • t h e n u m b e r of o u t c o m e s is 2
0 I ' / 'P '
'
4
3. T h e t o t a l v o l u m e o f t h e c l e a n i n g c o m p o u n d is
v o l u m e o f c o n c e n t r a t e a n d v o l u m e of w a t e r : 2 + 5.
T h e ratio o f v o l u m e o f c o n c e n t r a t e t o total v o l u m e of
t h e c l e a n i n g c o m p o u n d is 2 ; 7.
c ) v ) T h e value
- '
w r i t t e n a s a ratio, is 3 : 10,
d) iii) A n a r r a n g e m e n t o f o b j e c t s in a d e f i n i t e o r d e r is
a p e r m u t a t i o n of t h e o b j e c t s ,
e ) vii) If t h e p r o d u c t 5 - 4 3 2 1 is w r i t t e n a s 5!, it is
w r i t t e n in f a c t o r i a l n o t a t i o n ,
f) iv) A g r o u p o f o b j e c t s in w h i c h o r d e r d o e s not
m a t t e r is a c o m b i n a t i o n o f t h e o b j e c t s .
g) i) T h e s e t o f all p o s s i b l e o u t c o m e s is c a l l e d t h e
h) v i j ^ B r e n d a t o s s e d a c o i n 10 t i m e s . It t u r n e d u p tails
8 times. She used these results to determine that the
4. T h e r e a r e t e n l e t t e r s in S T A T I S T I C S .
S is r e p e a t e d t h r e e t i m e s , T is r e p e a t e d t h r e e t i m e s ,
a n d I IS r e p e a t e d t w i c e . T h e l e t t e r s c a n b e a r r a n g e d
10! w a y s . T o eliminate a r r a n g e m e n t s that w o u l d be
t h e s a m e b e c a u s e o f t h e r e p e a t i n g l e t t e r s , d i v i d e by
3!, 3! a n d 2 ! .
10!
^10-9-8-7-6-5-4-3!
3!-3!-2!^
10!
3I-3-2-1-2-1
^10-9-8-7-6-5-4
3!-3l-2!^
10!
3! 31
probaWlity Of t o s s i n g tails is A, T N s , s e n e x a m p l e
of experimental probability,
i) viii) S e r g e s a y s t h a t t h e p r o b a b i l i t y o f t o s s i n g h e a d s
w i t h a fair c o i n is ^ , b e c a u s e t h e l i k e l i h o o d t h a t t h e
2
c o i n will l a n d h e a d s is e g u a l to t h e l i k e l i h o o d t h a t it
will l a n d tails. T h i s is a n e x a m p l e of t h e o r e t i c a l
probability.
3 2 2
604 800
2!
12
10!
= 50 400
2!
T h e letters c a n b e a r r a n g e d in 5 0 4 0 0 w a y s .
3! 3!
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
5-41
b) T h e r e a r e f o u r w a y s t o roll 12 a n d 3 6 p o s s i b l e
o u t c o m e s . T h e p r o b a b i l i t y o f rolling a p r o d u c t o f 12 is
8!
2. a)
(8-5)!
4
1
— o r —.
36
9
8!
3!
•7 6 - 5 4 3 !
3!
HHH
7 6 5 4
HHT
,P5-6720
HTH
18!
b)
'
(18-2)!
HTT
THH
P = ^
-jg,
18^2
THT
18 17 1 6 !
^«
TTH
16!
= 18-1
17
2 "
is^2
T
306
P—ii"
TTT
b) T h e r e a r e t h r e e w a y s t o t o s s o n e h e a d a n d t w o
3
tails, a n d e i g h t p o s s i b l e o u t c o m e s : P ( 1 H, 2 T ) = — .
8
(4-4)!
4!
5. a) P u
0!
Q
4-3-2 1
. Q
4^4-4 3 2
/
.P
I
=24
1^8
=
1
i
11!
d)
\
J
'\
\
( 1 1 - 8 ) ! 8!
j
/
„
11!
3! - 8!
b)Pr^Q
11-10-9-8!
,a =
3-2-18!
P
/
11 10 9
11^8
'
.Co
=
/
\
/
3 2
990
C
\
-
\
6
Q
\
1
/
/
s /
X.
=165
e) i i C 3 = n C a . s i n c e 3 + 8 = 1 1
i i C 3 = 165
f) 24C24 = 1, s i n c e t h e r e is o n l y o n e w a y t o c h o o s e 2 4
from 24.
3. a)
P r o d u c t s of t h e R o l l s of T w o D i c e
D i e 1/
Die 2
1
2
3
4
5
6
6
1
1
2
3
4
5
2
2
4
6
8
10
12
3
3
6
9
12
15
18
4
4
8
12
16
20
24
5
5
10
15
20
25
30
6
6
12
18
24
30
36
5-42
C h a p t e r 5: P r o b a b i l i t y
d)P\QuQ\P
(8-4)!4!
8!
4! - 4!
8
7-6-5-4!
4!
4-3-2-1
8 7
6-5
4 3 2
1680
24
6. a) e . g . ,
Train
B
7:13
7:14
7:15
7:16
7:17
,C. = 70
Train A
7:11
7:12
7:13
7:14
7:15
7:16
11/13
11/14
11/15
11/16
11/17
12/13
12/14
12/15
12/16
12/17
13/13
13/14
13/15
13/16
13/17
14/13
14/14
14/15
14/16
14/17
15/13
15/14
15/15
15/16
15/17
16/13
16/14
16/15
16/16
16/17
b) i) T h e r e a r e f o u r w a y s t h e t r a i n s will a r h v e in t h e
There are 70 w a y s eight m e m b e r s can be c h o s e n to
chair four different committees.
1 0 . T h e r e is o n e w a y t o t o s s a 3, a n d six p o s s i b l e
o u t c o m e s . T h e p r o b a b i l i t y o f t o s s i n g a 3 o n c e is —.
6
s a m e minute, and there are 30 possible o u t c o m e s :
t h e p r o b a b i l i t y t h e t r a i n s will a r n v e in t h e s a m e m i n u t e
IS
4_
Using the Fundamental Counting Pnnciple:
6JUJ 3 6
30 •
ii) T h e r e a r e 2 0 w a y s t r a i n B will a r r i v e at l e a s t o n e
minute after train A has arnved, and there are
3 0 p o s s i b l e o u t c o m e s : t h e p r o b a b i l i t y t r a i n B will
a r n v e at l e a s t o n e m i n u t e a f t e r t r a i n A is
T h e p r o b a b i l i t y of t o s s i n g a 3 b o t h t i m e s w h e n a
6 - s i d e d die is t o s s e d t w i c e is 1
36
20
Chapter 5 Test, T R page
30 •
1 . T h e total n u m b e r of o u t c o m e s c a n b e d e t e r m i n e d
from the odds: 9 + 6 = 1 5 . From the o d d s against,
9 : 6, w e k n o w t h e n u m b e r o f w a y s t h e e v e n t c a n
7. S i n c e o r d e r d o e s not m a t t e r , this is a c o m b i n a t i o n
problem.
350
h a p p e n is 6. T h e r e f o r e , t h e p r o b a b i l i t y is — o r 0 4
15
12!
{l2-5)!5!
12!
12^5
7! • 5!
12H-10-9-8-7!
7! • 5 4 3 2
r
12 11-10 9 8
12^5
5-4-3-2
95 040
120
12C5
792
T h e r e a r e 7 9 2 w a y s J a s m i n e c a n invite 5 o f h e r
12 f n e n d s t o d n v e t o t h e ski resort.
8. T h e first digit m u s t b e a v o w e l , s o t h e r e a r e
5 o p t i o n s , t h e last m u s t a l s o b e a v o w e l , w i t h o u t
r e p e a t i n g t h e first s o t h e r e a r e 4 o p t i o n s .
5
4
F o r t h e m i d d l e t h r e e d i g i t s , t h e r e a r e 2 4 o p t i o n s left
for the s e c o n d , 23 for the third, and 22 for the fourth.
5, 2 4 , 2 3 , 2 2 , 4
By the Fundamental Counting Pnnciple,
5 - 2 4 - 2 3 - 2 2 - 4 or 2 4 2 8 8 0 f i v e - c h a r a c t e r
p a s s w o r d s can be f o r m e d under these conditions.
9. S i n c e o r d e r d o e s not m a t t e r , this is a c o m b i n a t i o n
problem.
F o u n d a t i o n s of M a t h e m a t i c s 12 S o l u t i o n s M a n u a l
2 . G e r m a i n e ' s p r o b a b i l i t y of w i n n i n g is
= -— or
^
8+ 5
13
a b o u t 0 . 6 2 . G a b r i e l ' s p r o b a b i l i t y o f w i n n i n g is
•z~— =
o r a b o u t 0 . 6 4 . G a b n e l s h o u l d start,
7+4
11
b e c a u s e his r e c o r d is b e t t e r
3. a) Let R r e p r e s e n t s e r v i n g h c e . Let C r e p r e s e n t
serving carrots.
P{R)
The
P{R
PiR
+ P ( C ) + P{R u cy = 0 . 7 8 + 0 . 3 0 + 0 . 1 4 = 1.22
total c a n n o t e x c e e d 1, s o P{R n C) m u s t b e 0 . 2 2 .
\ C) = 0 . 7 8 - 0 . 2 2 = 0.56
\ C) = 0 . 3 0 - 0 . 2 2 = 0 . 0 8
b) P{R n C ) = 0 . 2 2
5-43
4
: ;•• r u n bv: o'
d'.>'k r»f < ,1.(1% f .
of s e l e c t i n g 7 s p a d e s f r o m a
<
The n u m b e r of w a y s o f
S. a) T h i s is a c o n d i t i o n a l p r o b a b i l i t y p r o b l e m ,
P ( 2 r e d ) = P(1st r e d ) - P { 2 n d red | 1st red)
s«-lf,(.ii!ig , . n . / < .».o-. f r o m a d e c k o f c a r d s is 52C7.
1 if
P ( 2 r e d ) . f
f.
ol
/I
P ( 2 r e d ) = 0,318,..
')
r-; I ? 11
^'
r,
8
f, f> 4 J
1 7'
1 1 . - 11 K.. o a
f.
4 '. s
T h e p r o b a b i l i t y b o t h m a r b l e s a r e red is a b o u t 0 , 3 2 ,
b ) P ( 2 b l u e ) = P ( l s t b l n e ) - P { 2 n d blue 11st b l u e )
" >;d) f i ? ( ;
r,
P 2 blue)
/•pp
5
12 il 11 ,
i /1 n
52!
P ( 2 blue)
-^^^>!(52^7)i
52^7
T h e p r o b a b i l i t y b o t h m a r b l e s a r e b l u e is a b o u t 0 , 1 5 .
c ) P(1 red, 1 blue) = 1 ^ P(2 blue) - P(2 red)
P(1 red, 1 b l u e ) = 1 0.318.,. - 0 , 1 5 1 . , .
P(1 red, 1 blue) = 0,530...
T h e probability of drawing o n e red and o n e blue
m a r b l e is a b o u t 0 , 5 3 .
7, , 4 5 ,
5 2 ^ - 5 0 •49 - 48 - 47- 46 •45!
52-51-50-49-48-47-46
52 C ,
= 1 3 3 784 560
7. a) Let R r e p r e s e n t w a l k i n g b y t h e river, P ( P ) =
P ( 7 spades) =
52
0.151,
Let e r e p r e s e n t w a l k i n g a r o u n d t h e b l o c k . P ( B ) =
H
1716
P ( 7 spades) = - 3 3 ^ 3 ^ ^ g ^
T h e p r o b a b i l i t y of s e l e c t i n g 7 s p a d e s is
1716
5
-
j .
Let D represent seeing another d o g ,
P ( D / R) = 0 8 5
P ( D / B) = 0 , 3 0
133 7 8 4 5 6 0
another
PfR n D) = | | ) {,0,85} or 0.607 ,„
mmthm
f^B n O) = ( f ) {O.JO} or 0,0857 „,
dog
or a b o u t 0 . 0 0 0 0 1 2 8.
5. A s s u m e t h a t t h e 3 b o o k s a r e a r r a n g e d a s a g r o u p .
There are 7 other books, so I can arrange 8 objects
a s BPB or 8!. I c a n a l s o a r r a n g e t h e 3 b o o k s w i t h i n
t h e i r g r o u p a s 3!.
dog
T h e total n u m b e r o f w a y s to a r r a n g e all 10 b o o k s is
101.
,
V 8! • 3!
P ( t h e 3 books are together) =
,
.
X
8I-3-2-1
P ( t h e 3 books are together) =
b | P see another dog) =^p{Rn,D)
P see another dog) = P ( r ) - P ( d | r ) + P ( b ) - P ( o | b)
see another dog) =
P ( t h e 3 books are together) =
|
P ( t h e 3 books are together) =
^
P ( t h e 3 books are together) =
^
^
+ P{B n o )
5 1
12
.(0,85) + l -
-(O.SO)
p(see another dog) = 0,607,,, + 0,085.,,
p(see another dog) = 0.692.,,
T h e p r o b a b i l i t y t h a t L o g a n a n d his d o g will s e e
a n o t h e r d o g is a b o u t 0 . 6 9 .
T h e p r o b a b i l i t y t h a t t h e t h r e e b o o k s will b e p l a c e d
t o g e t h e r is ^
or a b o u t 0 . 0 7 .
15
5-44
C h a p t e r 5: P r o b a b i i i t f