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Probability Questions: A Level TYS Answers

probability questions for H2 A Level Statistics

Probability questions are found in paper 2, section B (Statistics section) of the H2 A Level Math exam. In this post, you will find the worked solutions for past A Level questions on probability.

A list of past questions from each statistics topic can be found here.

A quick summary of formulae in each statistics topics can be found here.

Past A Level Probability Questions

These are the past A Level probability questions for the new syllabus (syllabus 9758) which started from 2017.

Click on the link to go straight to the worked solutions for each question.

2017 Paper 2 Question 9 (vi) to (viii)

2018 Paper 2 Question 6

2018 Paper 2 Question 7

2019 Paper 2 Question 8

2020 Paper 2 Question 8

2021 Paper 2 Question 7c

Probability Question: 2017 Paper 2 Question 9 (vi) to (viii)

(vi)

probability tree diagram for probability and binomial distribution question of 2017 a level math paper 2 on statistics

P(not faulty light| identified as faulty)

= P(not faulty and identified as faulty) / P(identified as faulty)

= 0.92 (0.06) /[0.92 x 0.06 + 0.08 x 0.95]

= 0.42073

= 0.421 ( 3 s.f.)


(vii) P(quick test identifies correctly) = 0.08 (0.95) + 0.92(0.94) = 0.9408


(viii) For a light bulb that is identified as faulty, there is a 42.1% chance that it is not faulty. As this probability is relatively high, the quick test may not be worthwhile.

Probability Question: 2018 Paper 2 Question 6

(i) By observation, it’s observed that no matter which route is taken to reach D, it involves 8 steps, 3 of them are right fork and 5 of them are left fork.

Probability that bug finishes at D = 8! / (3!5!) p5q3 = 56p5q3 [shown]

(ii) Let X be the number of left forks made out of 8. X~B(8, p)

Given probability at D = P(X= 5) is maximum for this binomial distribution,

we know that P(X =5) > P(X =6) and P(X= 5) > P(X=4)

56p5(1-p)3 >8C6 p6(1-p)2

56p5(1-p)3 > 28p6(1-p)2

2(1-p)- p > 0

2-3p > 0

p<2/3

and

56p5(1-p)3 >8C4 p4(1-p)4

56p5(1-p)3 >70p4(1-p)4

4p – 5(1-p) > 0

p>5/9

Hence 5/9 <p <2/3

(iii) To reach any endpoint at A- I, bug will take 8 moves.

In each move, probability that bug is not swallowed by black hole = 1 – 0.1 = 0.9

To reach A-I without being swallowed by black hole, for all 8 moves, bug must not be swallowed by black hole.

Required probability = (0.9)8 = 0.43046721

Note: This question can be done as a binomial distribution. Part (ii) is testing students on mode of binomial distribution.

Probability Question: 2018 Paper 2 Question 7

This probability question is best down using either formulae and/ or sketching of Venn diagrams.

(i) Since A and B are independent, P(A∩B) = P(A) x P(B) = ab

P(A’∩B’) = 1 – P(AUB) = 1- [a + b – ab] = 1- a-b+ ab

P(A’) = 1- a; P(B’) = 1 -b

P(A’)xP(B’) = (1-a)(1-b) = 1-a-b+ab = P(A’∩B’)

Since P(A’) x P(B’) = P(A’∩B’), A’ and B’ are independent events. [shown]

(ii) Since A and C are mutually exclusive, P(A∩C) = 0

P(A’∩C’) = 1 – a – c

(iii) Given A’ and C’ are not mutually exclusive,

based on information from the first part, P(A) and P(B) are independent –> P(A)P(B) = P(A∩B)

max P(A∩B) = 2/15

min P(A∩B) = 1/3

Probability Question: 2019 Paper 2 Question 8

(i) (a) Required probability = (9+ 14)/56 = 23/ 56

(b) Required probability = (11+15)/56 = 13/ 28

(ii) (a) Required probability = 8/56 x 7/ 55 = 1/55

(b)

Case 1: yellow dog + not dog and not yellow

Probability for case 1 = 7/56 x (8+13+11)/55 x 2 = 8/55

Case 2: Non- yellow dog + yellow non dog

Probability for case 2 = 10/56 x 7/55 x 2 = 1/22

Total probability = 1/22 + 8/55 = 21/110

(iii) probability of selecting the 2 combination = 1/77 = (2 x 20)/(56 x 55)

20 = 5 x 4

Possible combinations are:

  • white horse and white rider
  • white horse and yellow bird
  • orange bird and yellow bird
  • orange bird and white rider

Probability Question: 2020 Paper 2 Question 8

(i)No. of ways to choose 12 shapes = 28C12 = 30 421 755

No. of ways to choose 1 rectangle and 11 circle = 17C1x 11C11= 17

No. of ways to choose 2 rectangles and 10 circles = 17C2x 11C10= 1496

P(R = 1) = 17/ 30 421 755

P(R = 2) = 1496/ 30 421 755 > P(R= 1)

Hence P(R= 1) < P(R=2)

(ii) 11 circles and (17+r) rectangles

Let total number of ways to choose 12 shapes from 11 circles and (17+r) rectangles be n.

No. of ways to choose 4 rectangles and 8 circles = 17+rC4 x 11C8= 165(17+rC4 )

No. of ways to choose 3 rectangles and 9 circles = 17+rC3 x 11C9 = 55(17+rC4 )

Given 165(17+rC4 )/ n = 15[ 55(17+rC3 )/ n]

17+rC4 = 5[17+rC3]

(17+r)!/ [4!(13+r)!] = 5[(17+r)!/(3!(14+r)!)]

1/ [4(13+r)!] = 5/ [(13+r)!(14+r)]

1/4 = 5/ (14+r)

r = 6

Probability Question: 2021 Paper 2 Question 7c

(a) No. of ways = 11! / (5! 2! 2!) = 83 160

(b) No fo ways = 4! (3)(2!) = 144

(c) Probability = [7!/ (2! 2!)] / 83 160 = 1/ 66

Note: part (a) and (b) are permutation and combination question, while part (c) is a permutation and combination cum probability question.

A Complete Course on H2 A Level Math Statistics

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