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Probability CA MS

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MEI AS Mathematics Probability

Topic assessment
1. In this question you must show detailed reasoning.
Which is more likely, never getting a 6 when rolling a fair die six times, or never getting a
double 6 when rolling a pair of fair dice 36 times? [4]

2. In the sixth form at Eastport School, there are 45 students studying at least one of the
three sciences, Biology, Chemistry and Physics.

The diagram below shows the number of students studying each science.

Biology Chemistry
12 15 2

4
3 6

3
Physics

(a) Find the probability that one of these students, selected at random, is studying
(i) Physics
(ii) Chemistry
(iii) both Physics and Chemistry. [3]

(b) Two students are selected at random. Find the probability that both are studying just
one science and that they are not both studying the same science. [4]

3. The probability distribution of a discrete random variable X is given by:

P( X = r ) = k (6r 2 − r 3 ) for r = 1, 2, 3, 4, 5
P( X = r ) = 0 otherwise
1
(a) Show that k = . [3]
105
Two values of X are chosen at random.

(b) Find the probability that the product of the two values is even. [3]

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MEI AS Maths Probability Assessment solutions

4. The Venn diagram shows two events, A and B, and their associated probabilities.

A B

p q 0.6

0.2

Given that events A and B are independent, find the values of p and q. [3]

5. The students in a school sixth form were asked for their most common mode of transport
to school. The replies are summarised in the table below.

Mode of transport
Walk/Cycle Public Private car Total
transport
Year 12 44 72 16 132
Year Group
Year 13 40 63 17 120
Total 84 135 33 252

(a) Two students are randomly selected. What is the probability that they are in different
year groups? [2]
(b) Two students are randomly selected. What is the probability that they are in different
year groups but are in the same ‘mode of transport’ category? [3]
(c) Based on the data given in the table, are the events ‘in Year 12’ and ‘walk/cycle to
school’ independent? Clearly explain your reasoning. [3]

6. A practical music examination can be taken once or twice. Those candidates who fail it
on the first occasion take it a second time.

For those having their first attempt, 25% pass with distinction and 45% gain an ordinary
pass. For those taking the examination for a second time, the corresponding figures are
5% and 70% respectively. The tree diagram below illustrates the situation.

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MEI AS Maths Probability Assessment solutions

First Second
attempt attempt
Pass with
Distinction

Ordinary Pass with


Pass Distinction

Ordinary
Fail Pass

Fail
(a) Find the probability that, following a second attempt if needed, a randomly chosen
candidate:
(i) fails the examination
(ii) passes the examination (with or without distinction) [4]

(b) Jill and Jo are two randomly chosen entrants for the examination. Find the
probabilities that
(i) both pass (with or without distinction), but just one of them needs a
second attempt,
(ii) Jill gets a better result than Jo. [8]

Total 40 marks

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MEI AS Maths Probability Assessment solutions

Solutions to topic assessment


6
5
1. P(no 6s with one die in six rolls) =   = 0.335
6
36
 35 
P(no double 6s with two dice in 36 rolls) =   = 0.363
 36 
So no double 6s is more likely. [4]

2. (a) (i) Total number of studying Physics = 4+3+6+3 = 16


16
P(Physics) =
45

(ii) Total number studying Chemistry = 15+2+4+6 = 27


27 3
P(Chemistry) = =
45 5

(iii) Total number studying both Physics and Chemistry = 4+6 = 10


10 2
P(Physics and Chemistry) = =
45 9
[3]

3 2 6
(b) P(P, C) = P(C, P) =  =
45 44 1980
3 12 36
P(P, B) = P(B, P) =  =
45 44 1980
2 12 24
P(C, B) = P(B, C) =  =
45 44 1980
6 + 36 + 24 66 1
Probability = 2  = = (= 0.0667 to 3 s.f.)
1980 990 15
[4]

3. (a)
r 1 2 3 4 5
P(X = r) 5k 16k 27k 32k 25k

5k + 16k + 27k + 32k + 25k = 1


105k = 1
1
k=
105
[3]

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MEI AS Maths Probability Assessment solutions

(b) P(product is even) = 1 - P(both numbers are odd)


= 1 − (5k + 27k + 25k )
2

2
 57 
= 1−  
 105 
864
=
1225
( = 0.705 to 3 sig. figs.)
[3]

4. p + q + 0.6 + 0.2 = 1 so p + q = 0.2


Events are independent; using P ( A  B ) = P ( A) P ( B ) : q = ( p + q )( 0.6 + q )
q = 0.2 ( 0.6 + q )
5q = 0.6 + q
4q = 0.6
q = 0.15
 p = 0.2 − 0.15 = 0.05
[3]

132 120 880


5. (a) P(Y12  Y13) + P(Y13  Y12) = 2   = = 0.501 (3 s.f.) [2]
252 251 1757

44  40 72  63 16  17
(b) 2 + 2 + 2 = 0.208 (3 s.f.) [3]
252  251 252  251 252  251

132 84
(c) P(Y12) = , P(walk/cycle to school) =
252 252
44
P(Y 12 and walk/cycle to school) =
252
132 84 44
 =
252 252 252
Therefore the events are independent.
Alternatively show that
44 132
P(Y12 walk/cycle to school) = P(Y12) : =
84 252
[3]

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MEI AS Maths Probability Assessment solutions

6.

First Second
attempt attempt
Pass with
Distinction
0.25

Ordinary Pass with


0.45 Pass Distinction
0.05
0.3
Fail 0.7 Ordinary
Pass

0.25
Fail

(a) (i) P(FF) = 0.3  0.25 = 0.075 or 3


40

(ii) P(passes) = 1 − 0.075 = 0.925 or 37


40
[4]

(b) (i) P(passing on first attempt) = 0.7


P(passing on second attempt) = 0.3  0.75 = 0.225

P(Jill passes on first attempt and Jo passes on second attempt)


= 0.7  0.225 = 0.1575
P(Jill passes on second attempt and Jo passes on first attempt)
= 0.7  0.225 = 0.1575
P(both pass but just one needs a second attempt)
= 0.1575 + 0.1575
= 0.315 or 63
200

(ii) P(passes with distinction) = 0.25 + 0.3  0.05 = 0.265


P(passes with ordinary pass) = 0.925 – 0.265 = 0.66
P(fails) = 0.075

P(Jo fails and Jill passes) = 0.075  0.925 = 0.069375


P(Jo gets an ordinary pass and Jill passes with distinction)
= 0.66  0.265 = 0.1749
P(Jill gets a better result than Jo) = 0.069375 + 0.1749
= 0.244 (3 s.f.)
[8]

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