Chapter 7
Teaching Notes, Chapter 7: The introduction to probability in Section 7.1 and the basic concepts
in Section 7.2 on using probability with set operations will be more familiar to most students than the
later sections. Section 7.3 in particular relies heavily on the combinatorial material developed in
Chapter 6. The concepts of conditional probability and independence of events are developed in
Section 7.4. Throughout the chapter, but particularly in Section 7.1, it will be helpful to emphasize
that the outcomes of a particular experiment may not all be equally likely.
Exploratory Exercise Set 7.1
1. Answers will vary, but the cup should land on its side significantly more often than right side up or
upside down; the empirical probabilities should reflect this.
2. Again, answers will vary. It is most likely that the empirical probabilities in part (e) will differ
slightly from the theoretical probabilities assigned in part (b).
3. (a) Believe Jane. Dick is mistaken because he is assuming that the four events of getting 0, 1, 2,
or 3 heads are equally likely; they are not.
(b) Answers will vary.
(c) Most often the answer will be closer to Jane’s prediction of 37.5%.
4. (a) (1) {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(2) {0, 1, 2, 3}
(3) {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(4) {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
(5) {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36}
(6) {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
(b) The sample spaces in (1), (3), and (6) are uniform.
5. (a) (1) N; each element of S is an outcome.
(2) S  {o1 , o2 ,o N }
(3) P( S )  P({o1 , o2 ,, o N })  P(o1 )  P(o2 )    P(o N )
P(o1 )  P(o2 )    P(o N )
P( S )  P(o1 )  P(o2 )    P(o N )  N  P(o1 )
1
P(S) = 1, so N  P(o1 )  1 and P(o1 )  .
N
1
.
N
(b) If E is any event of S with n(E) = k elements, o1 , o2 ,, ok , then by Rule 3 for a finite
probability space, P( E )  P(o1 )  P(o2 )    P(ok ) . From part (a), each outcome in E has
1 1
1
k n( E )
1
probability
, so P( E )       
.
N
N
N
N
n( S )
N
6. Answers will vary. Responses might mention that the two possible outcomes, raining and not
raining are not equally likely.
Since all outcomes are equally likely, each outcome has probability
Exercise Set 7.1
1. (a) S = {0, 1, 2, 3, 4}
(b) S = {M, I, S, P}
(c) {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
131
132
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Chapter 7
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
(d) {1, 2, 3, 4, …} (This is an infinite sample space.)
(a) E = {0, 2, 4}
(b) E = {M, S, P}
(c) E = {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6)}
(d) E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(a) the event of getting heads an odd number of times = {1, 3}
(b) the event of drawing a vowel = {I}
(c) the event that the sum of the numbers landing face down is odd
= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5)}
(d) the event that it takes more than 10 rolls to get doubles = {11, 12, 13, …}
(a) P(M) = 1/11; P(I) = 4/11; P(S) = 4/11; P(P) = 2/11
(b) P(M) + P(S) + P(P) = 1/11 + 4/11 + 2/11 = 7/11
(c) P(I) = 4/11
There are 4! = 24 possible arrangements of the students (see Section 6.1). For only one of these
arrangements are the students lined up as indicated, so the probability is 1/24.
There are C(52, 5) = 2,598,960 ways to choose a hand of five cards from a deck of 52 cards.
(a) There are 26 red cards in the deck, so there are C(26, 5) ways to choose 5 red cards. The
probability is C ( 26,5) /C (52 ,5)  65,780 / 2 ,598,960  253 / 9996  0.02531
(b) There are C(13, 2) ways to choose two hearts and C(13, 3) ways to choose 3 spades. By the
Multiplication Principle there are C (13, 2)  C (13, 3) such hands, so the probability is
C (13,2)  C (13,3)/ C (52,5)  22,308/2,5 98,960  143/16,660  0.008583
(c) There are 4 aces and 4 jacks in the deck, so the probability is
C(4,2)  C (4,3) /C (52 ,5)  24 / 2,598,960  1/ 108 ,290  0.000009234
(d) First choose one card from each of the four suits; then choose one card from the remaining 48
cards in the deck. There are 13 choices for each suit, so the probability is
C (13,1) 4  C (48,1) / C (52,5)  28,561  48 / 2,598,960  2197 / 4165  0.5275
(a) P(1) = 105/600 = 7/40, P(2) = 87/600 = 29/200, P(3) = 82/600 = 41/300, P(4) = 65/600 =
13/120, P(5) = 125/600 = 5/24, P(6) = 136/600 = 17/75.
(b) P(2) + P(4) + P(6) = 288/600 or 48% of the time
(c) P(1) + P(2) + P(3) = 274/600 or approximately 46% of the time
There are a total of 2 8  256 possible sequences, all equally likely.
(a) There is only C(8, 0) = 1 sequence with no heads, so the probability is 1/256.
(b) There are C(8, 2) = 28 sequences with exactly two heads, so the probability is 28/256 = 7/64.
(c) Since there are C(8, 0) + C(8, 1) = 9 sequences with fewer than two heads, there must be 256
– 9 = 247 sequences with at least two heads, so the probability is 247/256.
(d) There is only one sequence with no heads (see part (a)), so there are 256 – 1 = 255 sequences
with at least one head, and the probability is 255/256.
(c), (d), (e), and (f) could not be probabilities since they are not numbers between 0 and 1.
The outcomes of the previous tosses have no effect. If it is a fair coin, the probability is 1/2.
(a) Sectors A and B each take up 1/4 of the circle, so the probability is 1/4 that the pointer will
stop in each of those sectors. Sector D takes up 1/3 of the circle, so the probability that the
pointer will stop in Sector D is 1/3. That leaves 1 – 1/4 – 1/4 – 1/3 = 1/6 of the circle to be
equally divided between Sectors C and E, by symmetry. Thus, the probability that the pointer
will stop in each of Sectors C and E is 1/12.
(b) P(vowel) = P(A) + P(E) = 1/4 + 1/12 = 1/3.
(c) P(C or D) = P(C) + P(D) = 1/12 + 1/3 = 5/12.
(a) P(red) = 5/12.
(b) Since there are 4 red marbles and 7 blue marbles remaining in the bag, the probability is 4/11.
Probability
133
(c) There are C(5, 2) = 10 ways to choose 2 of the 5 red marbles, and a total of C(12, 2) = 66
ways to choose any two marbles from the bag, so the probability is 10/66 = 5/33.
5 4
5
(d) The product of the first two probabilities is the third:
  . This should not be
12 11 33
surprising considering the Multiplication Principle, since the process of choosing two red
marbles in part (c) is broken down into two steps in (a) and (b).
13. There are n 2 equally likely outcomes, each corresponding to an ordered pair of the form (i, j)
where i and j are positive integers less than or equal to n.
1
(a) The only ordered pair with a sum of 2 is (1, 1), so the probability is 2 .
n
(b) There are n – 1 possible ordered pairs with a sum of n: (1, n – 1), (2, n – 2), …, (n – 2, 2), and
n 1
(n – 1, 1). The probability is 2 .
n
(c) There are n possible ordered pairs with a sum of n + 1: (1, n), (2, n – 1), …, (n – 1, 2), and (n,
n 1
1). The probability is 2  .
n
n
1
(d) The only ordered pair with a sum of 2n is (n, n), so the probability is 2 .
n
Exploratory Exercise Set 7.2
1. (a) P( F  G ) , P( E  ( F  G))
(b) P( F  G)  P( F )  P(G)  P( F  G)
(c) E  G
(d) P( E  G ) , P(( E  F )  ( E  G))
(e) Using the commutative and associative properties of set intersection (see Section 2.1),
( E  F )  ( E  G)  ( E  E )  F  G  E  F  G .
(f) P( E  ( F  G))  P(( E  F )  ( E  G))  P( E  F )  P( E  G)  P( E  F  G)
(g) Putting it all together,
P ( E  F  G )  P ( E )  P ( F  G )  P ( E  ( F  G ))
 P ( E )  P ( F )  P (G )  P ( F  G )
 P ( E  F )  P ( E  G )  P ( E  F  G ) 
 P ( E )  P ( F )  P (G )
 P( E  F )  P( E  G )  P( F  G )
 P( E  F  G )
2. (a) P( A  B  C )  P( A)  P( B)  P(C )  P( A  B)  P( A  C )  P( B  C )  P( A  B  C )
= 0.50 + 0.30 + 0.45 – 0.15 – 0.25 – 0.20 + 0.10 = 0.75 or 75%
134
Chapter 7
(b)
A
B
.05
.2
.05
.1
.15
.1
.1
C
.25
(c) From the Venn diagram the probability is 0.25 or 1/4 or 25%.
3. Answers will vary. Emphasize that it is not always the case that P( A  B)  P( A)  P( B) .
Exercise Set 7.2
1. (a) P(scratched or cracked) = P(scratched) + P(cracked) – P(scratched and cracked)
= 0.10 + 0.12 – 0.05 = 0.17 or 17%
(b) This is the complement of the event in part (a), so the probability is 1 – 0.17 = 0.83.
2. (a) Since there are 12 face cards in the deck, the probability is 12/52 = 3/13.
(b) P( F  H )  P( F )  P( H )  P( F  H ) = 12/52 + 13/52 – 3/52 = 22/52 = 11/26.
(c) P( F  B )  1  P( F  B)  1  P( F )  P( B)  P( F  B) = 1 – [12/52 + 26/52 – 6/52]
= 1 – 32/52 = 1 – 8/13 = 5/13.
C (11, 3) 165
3. (a) P(3 heads) =

2048
211
 C (11,0)  C (11,1)  C (11,2) 
(b) P(at least 3 heads) = 1 – P(less than 3 heads) = 1  

211


1  11  55
67
1981
= 1
 1

11
2048 2048
2
4. (a) P(E ) = 1 – P(E) = 1 – (0.10 + 0.15 + 0.20) = 1 – 0.45 = 0.55
(b) P( E  F )  P( E )  P( F )  P( E  F ) = 0.45 + 0.60 – 0.10 = 0.95
(c) P( E  F )  1  P( E  F ) = 1 – 0.95 = 0.05
5. The number of three digit positive integers is 9  10  10 = 900.
(a) There are 8  9  9 = 648 three-digit positive integers with no 8s as digits, so the probability is
648/900 = 18/25.
(b) This is the complementary event of that in part (a) so the probability is 1 – 18/25 = 7/25.
(c) There are 7  8  8 = 448 three-digit positive integers with no digit greater than 7, so the
probability that a randomly chosen three-digit positive integer has at least one digit greater
than 7 is 1 – 448/900 = 452/900 = 113/225.
6. (a) Let A be the event that the number chosen is a multiple of 5, and let B be the event that the
number is a multiple of 7. Then P(A) = INT(100/5) / 100 = 20/100 = 1/5, P(B) = INT(100/7)
/ 100 = 14/100 = 7/50, and P( A  B) = INT(100/35) / 100 = 2/100 = 1/50, so
P( A  B)  P( A)  P( B)  P( A  B) = 20/100 + 14/100 – 2/100 = 32/100 = 8/25.
(b) P( A  B )  1  P( A  B) = 1 – 8/25 = 17/25.
7. Let M be the set of students taking math and let H be the set of students taking history.
(a) P( M  H )  P( M )  P( H )  P( M  H ) = 90/150 + 75/150 – 40/150 = 125/150 = 5/6.
Probability
135
(b) P( M  H )  1  P( M  H ) = 1 – 5/6 = 1/6.
8. Let A and B be the set of students who ate a sandwich and a salad, respectively.
(a) P( A  B)  P( A)  P( B)  P( A  B) = 0.5 + 0.6 – 0.2 = 0.9 or 90%.
(b) P( A  B )  1  P( A  B) = 1 – 0.9 = 0.1 or 10%.
9. We let B, C, and D represent the sets of students in the band, chorus, and drama club,
respectively.
(a) Using the formula from Exploratory Exercise 1 of this section,
P( B  C  D)  P( B)  P(C )  P( D)  P( B  C )  P( B  D)  P(C  D)  P( B  C  D)
= 150/300 + 180/300 + 165/300 – 70/300 – 85/300 – 90/300 + 40/300
= 290/300 = 29/30
(b) P( B  D  C )  P( B  D)  P( B  D  C) = 85/300 – 40/300 = 45/300 = 3/20 = 0.15
(c) P( B  D  C )  1  P( B  C  D) = 1 – 29/30 = 1/30.
10. We let A, B, and C represent the sets of campers who took archery, basketball, and canoeing,
respectively.
(a) Using the formula from Exploratory Exercise 1 of this section,
P( A  B  C )  P( A)  P( B)  P(C )  P( A  B)  P( A  C )  P( B  C )  P( A  B  C )
= 60/150 + 70/150 + 85/150 – 25/150 – 20/150 – 40/150 + 15/150
= 145/150 = 29/30
(b) P( A  B  C )  P( A)  P( A  B)  P( A  C)  P( A  B  C)
= 60/150 – 25/150 – 20/150 + 15/150
= 30/150 = 1/5 = 0.2
(c) P( A  B  C )  1  P( A  B  C ) = 1 – 29/30 = 1/30.
Exploratory Exercise Set 7.3
1. (a) No; there are not 4 equally likely outcomes.
(b) No; the five outcomes Bo describes are not equally likely.
(c) The correct answer is 6/16 = 3/8 = 0.375. There are 2 4  16 possible sequences of heads and
tails. The number of those sequences with exactly 2 heads is C(4, 2) = 6. Thus the
probability of getting exactly 2 heads is 6/16 = 3/8 = 0.375.
(d) Since the probability is 0.375, and 20(0.375) = 7.5, you should expect to get exactly 2 heads
about 7 or 8 times in 20 repetitions.
(e) Answers will vary.
2. (a) (Straight flush) First choose a suit (4 choices). The lowest card in the straight can be any card
from 2 through 10. Next choose the lowest card in the straight (9 choices). There are
49
36
4  9  36 possible straight flushes, so the probability is

 0.00001385 .
C (52, 5) 2,598,960
(b) (4 of a kind) There are 13 choices for the denomination and only C(4, 4) = 1 way to choose
all four cards of that denomination. Then there are 48 choices for the fifth card. The
13  48
624
probability is

 0.0002401 .
C (52, 5) 2,598,960
(c) (Full house) First choose the denomination with 3 of a kind (4 choices); then choose 3 cards of
that denomination (C(4, 3) = 4 choices); then choose the denomination with 2 of a kind (12
choices); then choose 2 cards of that denomination (C(4, 2) = 6 choices). The probability is
13  C (4, 3) 12  C (4, 2) 13  4 12  6
3744


 0.001441 .
C (52, 5)
2,598,960 2,598,960
136
Chapter 7
(d) (Flush) First choose the suit (4 choices); then choose 5 of the 13 cards of that suit (C(13, 5) =
1287 choices). However, we do not include the 36 straight flushes that we counted in part
4  1287  36
5112

 0.001967 .
(a). Thus the probability is
C (52,5)
2,598,960
(e) (Straight) First choose the lowest denomination in the straight (9 choices, as in part (a)). Then
choose a suit for each of the 5 cards in the straight ( 4 5  1024 choices). However, we do not
include the 36 straight flushes that we counted in part (a). Thus the probability is
9  45  36
9180

 0.003532 .
C (52,5)
2,598,960
(f) (3 of a kind) First choose the denomination with 3 of a kind (13 choices); then choose 3 cards
of that denomination (C(4, 3) = 4 choices). The fourth and fifth cards must be of a different
denomination from the 3 of a kind, and from each other, so choose 2 of the remaining 12
denominations for them (C(12, 2) = 66 choices); then choose one card from each of those
13  4  66 16
54,912
denominations ( 4  4  16 choices) The probability is

 0.02113 .
C (52, 5)
2,598,960
(g) (2 pairs) First choose 2 of the 13 denominations for the pairs (C(13, 2) = 78 choices); then
choose 2 of the 4 cards for each denomination ( C (4,2) 2  6 2  36 choices); finally choose a
fifth card of a different denomination from the two pairs (44 choices). The probability is
78  36  44
123,552

 0.0474
C (52, 5)
2,598,960
(h) (1 pair) Choose a denomination for the pair (13 choices); then choose 2 of the 4 cards of that
denomination (C(4, 2) = 6 choices). Next choose 3 of the remaining 12 denominations for the
other three cards (C(12, 3) = 220 choices); then choose one of the 4 cards from each of those
13  6  220  64 1,098,240
3 denominations ( 4 3  64 choices). The probability is

 0.4226
C (52, 5)
2,598,960
(i) (junk) Since all of the hands described above are mutually exclusive, the probability of getting
none of them is 1 – the sum of their probabilities. The probability is
36  624  3744  5112  9180  54,912  123,552  1,098,240
1,295,400
1
 1
2,598,960
2,598,960
1,303,560

 0.5016
2,598,960
1 1 1 1 1
3. (a) (1) When k = 3 we get     , which matches the answer to 7 (a).
2! 3! 2 6 3
1 1 1 1 1 1
9 3
(2) When k = 4 we get     

 , which matches 7 (b).
2! 3! 4! 2 6 24 24 8
1 1 1 1 1 1 1 1 1
1
1
1
(3) When k = 7 we get        



2! 3! 4! 5! 6! 7! 2 6 24 120 720 5040
2520  840  210  42  7  1

5040
1854 103
, which matches the answer in 7 (c).


5040 280
(b)
k
3
4
5
6
7
Probability
0.3333
0.375
0.3667
0.3681
0.3679
(c) As k increases, the probability seems to approach a number between 0.3 and 0.4.
Probability
137
(d)
k
3
4
5
6
7
Probability
0.3333
0.375
0.3667
0.3681
0.3679
Reciprocal
3
2.6667
2.7273
2.7170
2.7184
The reciprocals of the probabilities approach Euler’s constant, e  2.7182818 ... (see the table
above).
4. Answers will vary.
Exercise Set 7.3
1. For only one of the 5! = 120 possible arrangements are the books in alphabetical order; if they are
arranged randomly, then the probability is 1/5! = 1/120.
2. There are a total of P(26,3)  P(10,3)  26  25  24 10  9  8  11,232,000 possible license plates.
(a) If the license plate has 3 consecutive digits in ascending order, the digits could be 012, 123,
234, 345, 456, 567, 678, or 789. There are still P(26, 3) = 15,600 possible arrangements of
P ( 26,3)  8
8
1


the letters, so the probability is
.
P ( 26,3)  P(10,3) 10  9  8 90
(b) The arrangement of the digits could be 210, 321, 432, 543, 654, 765, 876, or 987; the
probability is the same as in part (a): 1/90.
(c) If there are three consecutive letters in alphabetic order (ABC, BCD, …, XYZ) then the first
letter can be any letter from A through X, so there are 24 such arrangements. The probability
24  P(10,3)
24
1


is
.
P( 26,3)  P(10,3) 26  25  24 650
3. Recall that there are 9  10  10  10  9000 possible 4-digit positive integers.
(a) The first digit cannot be a 0 or a 5; the remaining digits can be any digit other than 5, so the
8  93
648
81
probability is


 0.648 .
3
1000 125
9 10
81
44
(b) This is the complement of the event in (a) so the probability is 1 

 0.352 .
125 125
94
93
729
(c) The probability is


 0.729 .
3
3
1000
9 10
10
729
271
(d) This is the complement of the event in (c) so the probability is 1 

 0.271 .
1000 1000
4. (a) There are 52 – 4 = 48 non-aces in the deck, so the probability is
C ( 48,5) 1,712 ,304

 0.6588 .
C (52,5) 2,598,960
(b) There are C(4, 2) = 6 ways to choose two aces, and C(48, 3) = 17,296 ways to choose 3 cards
6 17,296
103,776
that are not aces, so the probability is

 0.03993 .
C (52, 5) 2,598,960
(c) There are 52 – 13 = 39 non-spades in the deck, so the probability is
C (39,5)
575,757

 0.2215 .
C (52,5) 2,598,960
(d) There are C(13, 2) = 78 ways to choose two spades, and C(39, 3) = 9139 ways to choose 3
78  9139
712,842
cards that are not spades, so the probability is

 0.2743 .
C (52, 5) 2,598,960
138
Chapter 7
C (6 2) 15
.

64
26
(b) There are C(6, 0) + C(6, 1) = 1 + 6 = 7 sequences with fewer than 2 heads, so the remaining
5. (a) There are C(6, 2) = 15 sequences with exactly 2 heads; the probability is
64 – 7 = 57 sequences must have at least 2 heads. The probability is
57
.
64
(c) There are C(6, 4) + C(6, 5) + C(6, 6) = 15 + 6 + 1 = 22 sequences with at least 4 heads, so
22 11
more heads than tails. The probability is
.

64 32
6. (a) There are 52 – 13 = 39 non-spades in the deck, so the probability is
C (39, 13)
8,122,425,444

 0.01279
C (52, 13) 635,013,559,600
(b) First choose 7 of the 13 spades; then choose 6 cards from the 39 non-spades in the deck. The
C (13, 7)  C (39, 6) 1716  3,262,623
probability is

 0.01279 .
C (52, 13)
635,013,559,600
(c) The complementary event is getting a hand with 0 or 1 spades. The probability is
C (39, 13)  C (13, 1)  C (39, 12)
58,962,792,112 576,050,767,488
1
 1

 0.9071 .
C (52, 13)
635,013,559,600 635,013,559,600
7. (a) There is only 1 winning combination out of the C(46, 6) = 9,366,819 possible combinations,
1
 0.0000001068 .
so the probability is
9,366,819
(b) Choose 5 of the 6 winning numbers (C(6, 5) = 6 choices), then choose 1 non-winning number
6  40
240
(C(40, 1) = 40 choices). The probability is

 0.00002562 .
C (46, 6) 9,366,819
8. If there are n numbers to choose from, we need for 1/C(n, 5) < 1/C(46, 6), or for
C(n, 5) > C(46, 6) = 9,366,819. Experimentation shows that C(66, 5) = 8,936,928 < C(46, 6)
while C(67, 5) = 9,657,648 > C(46, 6). There would have to be at least 67 numbers to choose
from.
9. We use the Inclusion-Exclusion Principle (see Section 6.6) to determine how many positive
integers < 1000 are multiples of 2, 5, 7, or 11. Let A, B, C, and D be the sets of positive integers
that are multiples of 2, 5, 7, and 11, respectively. Then
A1  n( A)  n( B )  n(C )  n( D )  INT (1000 / 2)  INT (1000 / 5)  INT (1000 / 7)  INT (1000 / 11)
= 500 + 200 + 142 + 90 = 932,
A2  n( A  B )  n( A  C )  n( A  D )  n( B  C )  n( B  D )  n(C  D )
= INT(1000/10) + INT(1000/14) + INT(1000/22) + INT(1000/35) + INT(1000/55) +
INT(1000/77) = 100 + 71 + 45 + 28 + 18 + 12 = 274,
A3  n( A  B  C )  n( A  B  D )  n( A  C  D )  n( B  C  D )
= INT(1000/70) + INT(1000/110) + INT(1000/154) + INT(1000/385)
= 14 + 9 + 6 + 2 = 31, and
A4  n( A  B  C  D )  INT(1000/770) = 1,
688
so n( A  B  C  D) = 932 – 274 + 31 – 1 = 688, and the probability is
= 0.688.
1000
10. From Section 6.4 the total number of possible combinations (with repetition) is
C(7 + 5 – 1, 7) = C(11, 7) = 330. (Think of arrangements of 7 tally marks ( / ) and 4 column
dividers ( | ).)
Probability
139
(a) If there are at least three cherry-flavored jellybeans, then we have four more jellybeans to
select, with no additional restriction. There are C( 4 + 5 – 1, 4) = C(8, 4) = 70 such
70
7
combinations (4 tally marks and 4 column dividers), so the probability is

 0.2121 .
330 33
(b) We use the Inclusion-Exclusion Principle. Let A, B, C, D, and E be the sets of combinations
with at least 3 jellybeans of the flavors apple, banana, cherry, date, and eggplant,
respectively. From our work in part (a),
A1  n( A)  n( B )  n(C )  n( D )  n( E ) = 5 70 = 350.
If a combination is in two of the five sets, it must have at least three jellybeans of two
different flavors. That leaves only one jellybean (of any flavor) to be chosen to complete the
sample of seven jellybeans. That means that
A2  n( A  B )  n( A  C )    n( D  E )  C (5, 2)  5  10  5  50 .
Clearly no sample of seven jellybeans can have at least three jellybeans of three or more
flavors, so A3  A4  A5  0 , and n( A  B  C  D  E )  A1  A2 = 350 – 50 = 300. Thus
300 10
the probability of getting at least three of some flavor of jellybeans is

 0.9091 .
330 11
11. Recall our work with non-adjacency conditions in Section 6.4. Suppose five of the tosses come
up heads. Considering the arrangement __ T __ T __ T __ T __ T __ , we see that there are C(6,
5) = 6 ways to fill in five of the six blanks with H’s, so there are 6 possible sequences with five
H’s, five T’s, and no two consecutive H’s. Similarly, if four of the tosses come up heads, then
from the arrangement __ T __ T __ T__ T__ T__ T __, we see that there are 6 T’s, 6 + 1 = 7
blanks, and C(7, 4) = 35 ways to fill 4 of the blanks with H’s. Indeed, if n is any whole number
less than or equal to 5, then there are C(10 – n + 1, n) = C(11 – n, n) sequences with n H’s and no
two consecutive H’s. Notice that if n > 5 then it is impossible to have a sequence with no two
consecutive H’s. Thus, there are a total of
C(6, 5) + C(7, 4) + C(8, 3) + C(9, 2) + C(10, 1) + C(11, 0) = 6 + 35 + 56 + 36 + 10 + 1 = 144
possible sequences with no two consecutive heads. Since there are a total of 210  1024 possible
144
9
sequences, the probability is
= 0.140625.

1024 64
Exploratory Exercise Set 7.4
1. (a) Answers will vary.
(b) Answers will vary.
(c) Partner A should be more successful in guessing than Partner B.
n( E  F )
n({2})
1
n( E  F )
n({4,6})
2


(d) (1) P(E | F) =
(2) P( E | F ) 


n( F )
n({1,2,3}) 3
n( F )
n({4,5,6}) 3
n( E  F )
n({1,3})
2
n( E  F )
n({5})
1
(3) P( E | F ) 
(4) P( E | F ) 




n( F )
n({1,2,3}) 3
n( F )
n({4,5,6}) 3
Roll ( x ) 1
2
3
4
5
6
2. Answers will vary. One possibility is
p( x )
0.1 0.3 0.2 0.1 0.2 0.1
3. (a) Answers will vary.
(b) Answers will vary.
(c) Answers will vary, but A’s strategy should be more successful.
(d) With switching, P(A) = 1/2.
(b) Without switching, P(B) = 1/3.
140
Chapter 7
4. (a)
0.98
Pos1
0.02
D
0.003
0.02
0.997
0.98
Neg1
Pos 2
0.98
Neg2
Pos 2
0.02
Neg2
Pos1 0.01
Pos 2
0.99
Neg2
0.01
Pos 2
Neg1 0.99
Neg2
0.01
D
0.99
(b) P( D | Pos1  Pos2 ) 

P( D  Pos1  Pos2 )
P( D  Pos1  Pos2 )  P( D  Pos1  Pos2 )
0.003(0.98) 2
0.00294

 0.9672
2
2
0.0030397
0.003(0.98)  0.997 (0.01)
5. Answers will vary.
Exercise Set 7.4
1. A probability cannot be greater than 1.
2. Answers will vary, but possible questions are given below.
(a) If a female is selected, what is the probability that she has brown hair?
(b) What is the probability that the person selected is female, given that he or she has brown hair?
(c) What is the probability that the person selected is a female with brown hair?
(d) What is the probability that the person selected is either a female or has brown hair?
(e) If a male is selected, what is the probability that he has brown hair?
3. (a) P(E | F)
(b) P(E | F)
(c) P( F | E )
(d) P( E | F )
(e) P(F | E)
n( F  R ) 12
3


 0.1154
4. (a) P( F | R ) 
n( R )
104 26
n( R  F ) 12 6


 0.3529
(b) P( R | F ) 
n( F )
34 17
P( S  I ) 0.3 10


 0.4762
5. (a) P( S | I ) 
P( I )
0.63 21
P( I  S ) 0.3 2

  0.6667
(b) P( I | S ) 
P( S )
0.45 3
6. To show that (ii) implies (iii), we assume that P(F | E) = P(F), and show that
P( F  E ) P( E  F )

, so
P( E  F )  P( E )  P( F ) . By definition, P( F | E ) 
P( E )
P( E )
P( E  F )
 P ( F ) . Multiplying both sides by P(E) gives P( E  F )  P( E )  P( F ) .
P( E )
To show that (iii) implies (ii), we assume that P( E  F )  P( E )  P( F ) and show that
Probability
141
P(F | E) = P(F). Since E is assumed to have nonzero probability, we can divide both sides of the
P( E  F ) P( E )  P( F )

 P( F ) . Since
given equation by P(E) to get
P( E )
P( E )
P( E  F ) P( F  E )

 P( F | E ) , then P(F | E) = P(F).
P( E )
P( E )
7. Using Theorem 5, we check whether P( E  F )  P( E )  P( F ) .
n( E ) 12 1
n( F ) 6 1
1 1 1

 and P(F) =

 , so P( E )  P( F )    .
P(E) =
n( S ) 36 3
n( S ) 36 6
3 6 18
2
1
P( E  F ) = P({2, 1), (2, 4)} =
  P( E )  P( F ) . Thus E and F are independent.
36 18
8. Using Theorem 5, we check whether P( E  F )  P( E )  P( F ) .
n( E ) 9 1
n( F ) 6 1
1 1 1

 and P(F) =

 , so P( E )  P( F )   
P(E) =
.
n( S ) 36 4
n( S ) 36 6
4 6 24
2
1
1
P( E  F ) = P({3, 1), (3, 5)} =


 P( E )  P( F ) . Thus E and F are not independent.
36 18 24
1 12 3
9. (a)
W
P( B1 W )  

2 22 11
12/22
B1
1/2
10/22
1 10 5
Y
P( B1 Y )  

2 22 22
1 7
7
W
P( B2 W )  

2 20 40
1/2
7/20
B2
13/20
1 13 13
Y
P( B2 Y )  

2 20 40
3 7 197
(b) P(W )  P( B1  W )  P( B2  W )  

 0.4477
11 40 440
P( B1  W ) 113
3 440 120
 197  

(c) P( B1 | W ) 
P(W )
11 197 197
440
10. P( D | Pos) 
P( D  Pos)
(0.9)(0.01)
0.009


 0.08411
P( Pos)
(0.9)(0.01)  (0.1)(0.98) 0.107
142
Chapter 7
11.
Pos
0.98
P( D  Pos)  0.003  0.98  0.00294
D
0.003
0.02
Neg
Pos
0.01
0.997
P( D  Neg)  0.003  0.02  0.00006
P( D  Pos)  0.997  0.01  0.00997
D
0.99
Neg
12. P( D | Neg) 
P( D  Neg)  0.997  0.99  0.98703
P( D  Neg)
(0.003)(0.02 )
0.00006


 0.000061 .
P( Neg)
(0.003)(0.02 )  (0.997 )(0.99 ) 0.98709
13. (a)
Pos
0.995
P( D  Pos)  0.008  0.995  0.00796
D
0.008
0.005
Neg
Pos
0.02
0.992
P( D  Neg)  0.008  0.005  0.00004
P( D  Pos)  0.992  0.02  0.01984
D
0.98
Neg P( D  Neg)  0.992  0.98  0.97216
P( D  Pos)
0.008  0.995
0.00796


 0.2863
(b) P( D | Pos) 
P( Pos)
0.008  0.995  0.992  0.02 0.0278
(c) P( D | Neg) 
P( D  Neg)
0.992  0.98
0.97216


 0.99996
P( Neg)
0.008  0.005  0.992  0.98 0.9722