OCR Statistics 1 Probability
Section 1: Introducing probability
Solutions to Exercise
1. (i)
(ii)
P(5 ) 
1
8
P(5 or more) = P(5, 6, 7 or 8) 
4
8

1
2
(iii) P(less than 5) = 1 – P(5 or more)  1  21 
(iv)
2. (i)
(ii)
P(multiple of 2) = P(2, 4, 6 or 8) 
P(3) 
4
8

1
2
1
2
1
12
P(even number) = P(2, 4, 6, 8, 10 or 12) 
6
12

(iii) P(multiple of 3 and even number) = P(6, 12) 
1
2
2
12

1
6
(iv)
P(multiple of 3 or even number) = P(2, 3, 4, 6, 8, 9, 10 or 12) 
(v)
P(neither multiple of 3 or even number)  1  23 
8
12

2
3
1
3
3. P(A)  P(B)  0.4  0.3
 0.7
 P( A or B )
so A and B are not mutually exclusive.
P(B)  P(C)  0.3  0.25
 0.55
 P(B or C)
so B and C are mutually exclusive.
P(A)  P(C)  0.4  0.25
 0.65
 P(A or C)
so A and C are mutually exclusive.
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OCR S1 Probability 1 Exercise solutions
4.
1
2
3
4
5
6
7
8
1
0
1
2
3
4
5
6
7
Outcome
0
1
2
3
4
5
6
7
2
1
0
1
2
3
4
5
6
3
2
1
0
1
2
3
4
5
4
3
2
1
0
1
2
3
4
5
4
3
2
1
0
1
2
3
6
5
4
3
2
1
0
1
2
7
6
5
4
3
2
1
0
1
8
7
6
5
4
3
2
1
0
Probability
8
1
64  8
7
14
64  32
12
3
64  16
10
5
64  32
8
1
64  8
6
3
64  32
4
1
64  16
2
1
64  32
5.
0.3
0.7
late
E
0.3
late
0.7
not
0.3
late
not
0.7
not
(i) P(both on time)  0.7  0.7  0.49
(ii) P(one on time)  0.7  0.3  0.3  0.7  0.42
6. (i) P(train not late) = 0.9
P(train not late on four journeys)  0.94  0.6561
P(at least one train late) = 1 – 0.6561 = 0.3439
(ii) P(train not late on 10 journeys)  0.910  0.3487
P(at least one train late) = 1 – 0.3487 = 0.6513 (4 s.f.)
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OCR S1 Probability 1 Exercise solutions
7.
0.6
delay
0.4
not
delay
0.4
0.6
0.3
0.7
not
delay
0.3
not
delay
not
delay
0.7
delay
0.6
delay
0.7
0.3
not
0.7
not
0.4
0.3
not
(i) P(DDD)  0.4  0.6  0.7  0.168
(ii) P(delayed at one) = P(DNN) + P(NDN) + P(NND)
 (0.4  0.4  0.3)  (0.6  0.6  0.3) (0.6  0.4  0.7)
 0.324
(iii)P(delayed at none)  0.6  0.4  0.3  0.072
P(delayed at one) = 0.324 (from (ii))
P(delayed at two or more) = 1 – 0.072 – 0.324 = 0.604
8. There are 24 students who study Maths. Of these, 15 study English.
15 5
P(E | M ) 

24 8
30 3

50 5
15
3

15 students select History, so P(H ) 
50 10
9
3 3
P(E )  P(H )  

5 10 50
9 students select both English and History, so
9
P(E and H ) 
 P(E )  P(H )
50
so E and H are independent events.
9. 30 student select English, so P(E ) 
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OCR S1 Probability 1 Exercise solutions
10.
L
0.5
L
0.4
0.5
O
0.3
0.6
L
O
0.7
O
(i) P(exactly one journey is on time)  (0.4  0.5 )  (0.6  0.3)
 0.2  0.18
 0.38
(ii) Let A be the event that the first journey is on time
Let B be the event that the second journey is on time
P( A and B )  0.6  0.7  0.42
P(B )  0.4  0.5  0.6  0.7  0.2  0.42  0.62
P( A and B ) 0.42
P( A |B ) 

 0.677 (3 s.f.)
P(B )
0.62
11.
0.6
clears
E
0.75
0.4
clears
fails
0.25
0.3
clears
0.7
fails
fails
(i) P(fails on all 3 attempts)  0.4  0.25  0.7  0.07
P(clears height)  1  0.07  0.93
(ii) Let A be the event that she clears the height
Let B be the event that she clears the height on the first attempt
P( A and B )  0.6
P(B )  0.93 (from part (i)
P( A and B ) 0.6
P(B | A ) 

 0.645 (3 s.f.)
P( A )
0.93
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