Probability Revision – This work should take you no longer than 40 minutes – Calculator allowed
1.
2.
Two unbiased 6-sided dice are rolled, a red one and a black one. Let E and F be the events
E : the same number appears on both dice;
F : the sum of the numbers is 10.
Find
(a)
P(E);
(b)
P(F);
(c)
P(E  F).
(Total 6 marks)
L
Dumisani is a student at IB World College.
The probability that he will be woken by his alarm clock is
7
.
8
If he is woken by his alarm clock the probability he will be late for school is
1
.
4
W
If he is not woken by his alarm clock the probability he will be late for school is
3
.
5
L
L
Let W be the event “Dumisani is woken by his alarm clock”.
Let L be the event “Dumisani is late for school”.
(a)
Copy and complete the tree diagram on the right.
(b)
Calculate the probability that Dumisani will be late for school.
(c)
(3)
Given that Dumisani is late for school what is the probability that he was
woken by his alarm clock?
W
(4)
L
(Total 11 marks)
(4)
In a school of 88 boys, 32 study economics (E), 28 study history (H)
and 39 do not study either subject. This information is represented in
the following Venn diagram.
3.
E(32)
H(28)
(a) Calculate the values a, b, c.
a
b
(4)
c
(b) A student is selected at random.
(i) Calculate the probability that he studies both economics and
history.
(ii) Given that he studies economics, calculate the probability that he
does not study history.
(3)
(c)
A group of three students is selected at random from the school.
(i)
Calculate the probability that none of these students studies economics.
(ii)
Calculate the probability that at least one of these students studies economics.
(5)
(Total 12 marks)
4.
A fair coin is tossed eight times. Calculate
(a)
the probability of obtaining exactly 4 heads;
(2)
(b)
the probability of obtaining exactly 3 heads;
(c)
the probability of obtaining 3, 4 or 5 heads.
(1)
(3)
(Total 6 marks)
5.
For events A and B, the probabilities are
P ( A) 
3
4
, P( B)  .
11
11
1
ANSWERS
1.
Total number of possible outcomes = 36 (may be seen anywhere)
(a)
P( E)  P(1,1)  P(2, 2)  P(3, 3)  P(4, 4)  P(5, 5)  P(6, 6)
6

36
(b)
P( F )  P(6, 4)  P(5, 5)  P(4, 6)
3

36
P  E  F   P( E )  P( F )  P( E  F )
(c)
1
36
6
3 1  8 2

P E  F    
 , 0.222 

36 36 36  36 9

P( E  F ) 
(A1)
(A1) (C2)
(A1) (C1)
(A1)
(M1)(A1) (C3)
[6]
2.
(a)
1
4
7
8
W
3
4
3
5
1
8
(c)
L
L
W
2
5
(b)
L
L
(A1)(A1)(A1)(A1)
4
 7 1 3
Note: Award (A1) for the given probabilities  , ,  in the correct positions,
8 4 5
and (A1) for each bold value.
7 1 1 3
Probability that Dumisani will be late is   
(A1)(A1)
8 4 8 5
47
=
(0.294)
(A1)
3
160
P(W  L)
P(WL) =
P(L)
7 1
P(W  L) = 
(A1)
8 4
47
P(L) =
(A1)
160
7
P(WL) = 32
(M1)
47
160
35
=
(= 0.745)
(A1)
4
47
2
3.
(a)
U(88)
E(32)
H(28)
a
c
b
39
n (E  H) = a + b + c = 88 – 39 = 49
n (E  H) = 32 + 28 – b = 49
60–49 = b = 11
a = 32 – 11 = 21
c = 28 – 11 = 17
Note: Award (A3) for correct answers with no working.
(b)
(i)
(ii)
P(E  H) =
11 1

88 8
(M1)
(A1)
(A1)
(A1)
4
(A1)
21
PH ' E  88

P(HE) =
32
P E 
88
21
=
(= 0.656)
32
(M1)
(A1)
OR
21
32
56  55  54
P(none in economics) =
88  87  86
Required probability =
(c)
(i)
(A1)(A1)
3
(M1)(A1)
= 0.253
(A1)
3
Note:
Award no marks for
(ii)
 56 
 = 0.258.
 88 
Award (M0)(A0)(A1)(ft) for 
56  55  54
.
88  88  88
P(at least one) = 1 – 0.253
= 0.747
OR
 32 56 55   32 31 56  32 31 30
    3      
 88 87 86   88 87 86  88 87 86
3 
= 0.747
(M1)
(A1)
(M1)
(A1)
5
[12]
3
 8  1   1 
p(4 heads) =     
 4  2   2 
4
4.
(a)
8 4
(M1)
8
8 7 65  1 
 
1 2  3  4  2 
70
=
 0.273 (3 s.f.)
256
3
8 3
8
 8  1   1 
8 7 6  1 
p(3 heads) =      
 
1 2  3  2 
 3 2   2 
=
(b)
56
 0.219 (3 s.f.)
256
p(5 heads) = p(3 heads) (by symmetry)
p(3 or 4 or 5 heads) = p(4) + 2p(3)
70  2  56 182
=

256
256
 0.711 (3 s.f.)
=
(c)
(A1)
2
(A1)
1
(M1)
(M1)
(A1)
3
[6]
5.
4