Discrete random variable 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 discrete random variable.

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 Discrete Random Variable Questions

These are the past A Level discrete random variable 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.

## Discrete Random Variable Question (DRV): 2017 Paper 2 Question 5

(i) P(T=2) = (6/9)(5/8) = 5/12

P(T=3) = (6/9)(5/8)(3/7)(2) = 5/14

P(T=4) = (6/9)(5/8)(3/7)(2/6)(3)= 5/28

P(T=5) = (6/9)(5/8)(3/7)(2/6)(1/5)(4)= 1/21

(ii) E(T) = 5/12(2) + 5/14(3) + (5/28)(4)+(1/21)(5) = 20/7

E(T^{2}) = 5/12(2^{2}) + 5/14(3^{2}) + (5/28)(4^{2})+(1/21)(5^{2}) = 125/14

Var(T) = 125/14 – (20/7)^{2} = 75/98

(iii) P(T≥4) = 5/28 + 1/21 = 19/84

Let X be the number of games played out of 15 where Lee takes at least 5 counters out of the bag in each game.

X~B(15, 19/84)

P(X≥5) =1 -P(X≤4) = 0.238 ( 3 s.f.)

## Discrete Random Variable Question (DRV): 2018 Paper 2 Question 8

(i) two 3, three 4, n 5.

Possible values of S:

ball 1 /ball 2 | 3 | 4 | 5 |

3 | 6 | 7 | 8 |

4 | 7 | 8 | 9 |

5 | 8 | 9 | 10 |

S | 6 | 7 | 8 | 9 | 10 |

P(S=s) | 2/ [(n+4)(n+5)] | 12/ [(n+4)(n+5)] | (6+4n)/[(n+4)(n+5)] | 6n/[(n+4)(n+5)] | n(n-1)/[(n+4)(n+5)] |

(ii) when n = 1, P(S=10) = 0

There is no possibility that S = 10. In order for S=10, two balls numbered 5 have to be picked. However, since there is only 1 ball numbered 5, it is not possible to pick 2 balls numbered 5 (without replacement).

(iii) E(S) = 12/ [(n+4)(n+5)] +84/ [(n+4)(n+5)] +(48+32n)/[(n+4)(n+5)] + 54n/[(n+4)(n+5)] +10n/(n-1)/[(n+4)(n+5)] =(144+76n+10n^{2})/[(n+4)(n+5)] =2(n+4)(5n+18)/[(n+4)(n+5)] = (10n+36)/[(n+5)]

E(S^{2}) = 72/ [(n+4)(n+5)] +588/ [(n+4)(n+5)] +(384+256n)/[(n+4)(n+5)] + 486n/[(n+4)(n+5)] +100n/(n-1)/[(n+4)(n+5)] =(1044+642n+100n^{2})/[(n+4)(n+5)]

Var(S) =E(S^{2}) – [E(S)]^{2} = (1044+642n+100n^{2})/[(n+4)(n+5)] – (10n+36)^{2}/[(n+5)^{2}]

= [(1044+642n+100n^{2})(n+5) -(100n^{2}+720n+1296)(n+4)]/ [(n+4)(n+5)^{2}]

= [22n^{2} + 78n + 36]/ [(n+4)(n+5)^{2}]

g(n) = 22n^{2} + 78n + 36

## Discrete Random Variable Question (DRV): 2020 Paper 2 Question 5

(i) 1 green; r red; 2r blue; Total number of discs = 3r+1

green – 0 points; red – 5 points; blue – 2 points

possible scores

green (0) | red (5) | blue (2) | |

green (0) | N/A [not possible, only 1 green] | 0 | 0 |

red (5) | 0 | 25 | 10 |

blue (2) | 0 | 10 | 4 |

Tina’s possible scores are: 0, 4, 10, 25

(ii) Let Tina’s score be S.

S | 0 | 4 | 10 | 25 |

P(S=s) | 2(3r)/[3r(3r+1)] = 2/(3r+1) | 2r(2r-1)/[3r(3r+1)] =(4r-2)/[3(3r+1)] | 4r^{2}/[3r(3r+1)]=4r/[3(3r+1)] | r(r-1)/[3r(3r+1)] =(r-1)/[3(3r+1)] |

E(S) = 0+ (16r-8)/[3(3r+1)] +40r/[3(3r+1)] + (25r-25)/[3(3r+1)] =(81r-33)/[3(3r+1)]=(27r-11)/(3r+1)

E(S^{2}) = (64r-32)/[3(3r+1)] + 400r/[3(3r+1)] +(625r-625)/[3(3r+1)] =(1089r – 657)/ [3(3r+1)] = (363r-219)/(3r+1)

Var(S) =(363r-219)/(3r+1) – [ (27r-11)/(3r+1) ] ^{2} = [(363r-219)(3r+1) – (27r-11)^{2}]/ (3r+1)^{2} = (360r^{2} + 300r-340 )/(3r+1)^{2}

(iii)when Var(S) = 38,

(360r^{2} + 300r-340 )/(3r+1)^{2} = 38

Using Graphic calculator, and rejecting negative values of r,

r = 3

## Discrete Random Variable Question (DRV): 2021 Paper 2 Question 6

Given:

score | 1 | 2 | 3 | 4 | 5 |

probability | 0.2 | 0.3 | p | p | q |

Also given variance = 1.61

0.2 + 0.3 + p + p + q = 1

2p + q = 0.5

q= 0.5 – 2p —-(1)

Let S be the score.

E(S) = 0.2 + 0.6 + 3p + 4p + 5q = 0.8 + 7p + 5q sub (1) in

E(S) = 0.8 + 7p +5(0.5-2p) = 3.3 – 3p

E(S^{2}) = 0.2 + 1.2 + 9p + 16p + 25q = 1.4 + 25p + 25q sub (1) in

E(S^{2}) = 1.4 + 25p + 25(0.5 – 2p) = 13.9 – 25p

Var(S) = E(S^{2}) -[E(S^{2})] = 13.9 – 25p -(3.3-3p)^{2} = 13.9 – 25p – 10.89 +19.8p – 9p^{2} = 3.01-5.2p – 9p^{2}

3.01 – 5.2p- 9p^{2} =1.61

By Graphic Calculator, and rejecting negative values

p = 0.2

E(S) = 3.3 – 3(0.2) = 2.7

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