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