2023 Stats3 Discrete Random Variables APQ solutions (RVHS)

Statistics 3 Tutorial: Discrete Random Variables Additional Practice Questions 1. An unbiased cubical dice has three faces numbered ‘1’, two faces numbered ‘2’ and one face numbered ‘3’. The random variable X is the number showing on the top face of the dice when it is thrown. Show that 5E( ) 3X = and find Var( )X . Solution: X = x 1 2 3 P(X = x) 3 6 2 6 1 6 2 2 2 2 2 3 4 3 10 5E( ) 6 6 6 6 3 3 2 1 20E( ) 1 2 3 6 6 6 6 20 5 5Var( ) 6 3 9 X X X = + + = = = + + = = − = 2. [2001/II/6] A random variable X has the probability distribution given in the following table. x 2 3 4 5 P(X = x) p 2 10 3 10 q (i) Given that E(X) = 4, find p and q. (ii) Show that Var(X) = 1. (iii) Find E(|X – 4|). (iv) Ten independent observations of X are taken. Find the probability that the value 3 is obtained at most three times. Solutions: (i) 23E( ) 4 2 3 4 5 4 10 10 11 2 5 (1) 5 X p q pq    =  + + + =        + = −−− Also, 2 3 1 1 (2)10 10 2p q q p+ + + =  = − −−− Subs. (2) into (1): 1 1125 25pp + − = 12 10 5pq=  =

(ii) 2 2 2 2 2 1 2 3 2E( ) 2 3 4 5 1710 10 10 5X        = + + + =               2 2 2Var( ) E( ) E( ) 17 4 1X X X= − = − = (shown) (iii) 1 2 3 2 4E(| 4 |) | 2 | | 1| | 0 | |1|10 10 10 5 5X        − = − + − + + =               (iv) Let Y denote the no. of times (out of 10) ‘3’ is obtained. Then 2~ B 10, 5Y   Required probability = ( 3) 0.879PY = (to 3 sf) 3. [2017 PJC MYE/ P2/ Q7] In a game, 3 red balls and 7 white balls are placed in a bag. A player draws 3 balls at random and without replacement. The player scores 2 poi nts for every red b all that is drawn, and 1 point for every white ball that is drawn. The total score X is obtained by adding up the scores of the 3 balls. Find the probability distribution of X. [3] Find ( )E X and show that ( )Var 0.49X = . [2] x 3 (3W) 4 (2W1R) 5 (1W2R) 6 (3R) ( )P Xx= 7 6 5 7 10 9 8 24  = OR 7 3 10 3 7 24 C C = 7 6 3 3! 21 10 9 8 2! 40   = OR 73 21 10 3 21 40 CC C = 7 3 2 3! 7 10 9 8 2! 40   = OR 73 12 10 3 7 40 CC C = 3 2 1 1 10 9 8 120  = OR 3 3 10 3 1 120 C C = ( ) 7 21 7 1E 3 4 5 6 3.924 40 40 120X =  +  +  +  = ( ) 2 2 2 2 2 7 21 7 1E 3 4 5 6 15.724 40 40 120X =  +  +  +  = ( ) ( ) ( ) ( ) 2 22Var E E 15.7 3.9 0.49X X X= − = − = 4. [2017 SRJC MYE/ P2/ Q10 (part of)] (a) It is known that X is a discrete random variable such that ( ) ( ) 24E Var 3 2 1XX = + + . Find E(X). [2] (b) To promote the sales of a snack for children from March to August, one game card is inserted into each box of the snack. For each month, the game cards feature a different series of comic book characters and there are twelve distinct game cards in each series. Andy buys four boxes of the snack in March.

The random variable X denotes the number of distinct games cards that Andy will get from his purchases. (i) Show that P(X = 3) = 55 144 and P(X = 2) = 77 1728 . Hence, tabulate the probability distribution of X. [4] (ii) Find E(X) and Var(X). [2] (a) 4E(X2) = Var(3 + 2X) +1.  4E(X2) = 4Var(X) + 1 = 4E(X2) – 4[E(X)]2 + 1  4[E(X)]2 = 1  E(X) = 1 2 (b)(i) P(X = 3) = 1 11 10 12 12 12            + 11 2 10 12 12 12            + 11 10 3 12 12 12            OR 12 11 12 4 4!CC 2! 12  = 55 144 P(X = 2) = 12 2 4 4!C 2!2! 12  + 12 11 11 4 4!CC 3! 12  = 77 1728 X 1 2 3 4 P(X = x) 3 1 12   = 1 1728 OR: 12 1 4 C 12 77 1728 55 144 11 10 9 12 12 12            = 55 96 OR: 12 4 4 P 12 (ii) E(X) = 1 1728 + 772 1728   + 553 144   + 4 55 96   = 6095 1728 OR 3.527199 = 3.53 (to 3 s.f.) (By GC) E(X2) = 1 1728 + 2 772 1728   + 2 553 144   + 2 554 96   = 7363 576

Var(X) = 7363 576 – 2 6095 1728   OR Using GC, ( ) 2 0.5846817963 = 1020767 2985984 or 0.342 (3sf) 5. [2018 VJC H2 MYE Qn10] A board of directors consists n men and ( )10 n− women, where 37 n . A committee consisting of 3 randomly chosen people is to be formed. Let W be the number of women in the committee. (i) Show that ( ) ( )( )1 10P1 240 n n nW −−== and find , in terms of n, the probability distribution of W. [4] (ii) Given ( )E 2.1W = , find n. [2] (iii) Hence, find Var(W). [1] Solutions: i ( ) ( ) ( ) ( ) ( )( ) P 1 P P P 10 1 310 9 8 1 10 240 W MMW MWM WMM nnn n n n = = + + −−=    −−= Alternative Method: ( ) ( ) ( )( ) 10 21 10 3 1 10 2! 1!P1 120 1 10 240 nn nn n CCW C n n n − − −= = = −−= w ( )P Ww= 0 ( )( ) 312 or 720 120 nn n n C−−   1 ( )( )1 10 240 n n n−− 2 ( )( ) ( ) 10 210 9 or 240 120 nnCn n n −−−  3 ( )( )( ) 10 310 9 8 or 720 120 nn n n C−− − −  

ii ( ) ( )( ) ( )( ) ( )( )( ) 1 10 10 9E2 240 240 10 9 83 720 n n n n n nW n n n − − − − =+   − − −+   ( ) ( ) ( )( )( ) ( ) ( ) ( )( )( ) ( ) ( ) 222 102.1 3 1 6 9 3 9 8720 10 1 2 9 9 8240 10 18 2 72 17240 0.3 10 0.3 0.9 3 n n n n n n n n n n n n n n n n n n n n n n n n −= − + − + − − −= − + − + − − −= − + − + − + =− = = Or By GC, n = 3 iii By GC, Var(W) = 0.49 w 0 1 2 3 ( )P Ww= 1 120 7 40 21 40 7 24 6 [2018 SRJC H2 MYE P2 Qn8] An unbiased 4-sided die has its faces numbered 1, 2, 3 and 4. When the die is thrown, the number on the face in contact with the table is noted. Two such dice are thrown and the score X is found by multiplying these numbers together. (i) Obtain the probability distribution of X. [2] (ii) Using an algebraic method, show that the value of ( )E5 X − is 27 8 . [1] (iii) Two independent observations of X are taken. Find the probability that one of them is 2 and the other is at most 3. [2] Solution (i) Table of outcomes (Not a required working, but good to have) Die 1 Die 2 1 2 3 4 1 1 2 3 4 2 2 4 6 8 3 3 6 9 12 4 4 8 12 16 Probability distribution of X:

7 [2018 HCI J2 MYE Q12] A biased 6-sided die is numbered from 1 to 6. The number obtained when the die is thrown is denoted by X. The probability distribution of X is shown in the following table: Number shown on die, x 1 2 3 4 5 6 Probability ( )P Xx= 0.1 a 0.2 0.2 b 0.1 where a > 0 and b > 0. (i) Write down an equation satisfied by both a and b. [1] (ii) If E( ) 3.5X = , what can you say about the relationship between a and b? [1] It is given that a > b, (iii) Find E( )X in terms of b. Using the result in (i), find an inequality in the form E( )X , where and are constants to be determined. [3] (iv) Find Var( )X in terms of b. [3] The biased die is thrown two times, the first number obtained is denoted as 1X and the second number obtained is denoted as 2X . (v) Find the value of 12P( 4)XX− . [3] x 1 2 3 4 6 8 9 12 16 P(X=x) 1 16 1 8 1 8 3 16 1 8 1 8 1 16 1 8 1 16 – Correct values for all outcomes – Correct values for all probabilities (ii) 5x− 1 2 3 4 7 11 ( )P5 xx−= 5 16 1 8 1 4 1 8 1 8 1 16 ( ) 5 1 1 1 1 1E 5 1 2 3 4 7 1116 8 4 8 8 16x            − = + + + + +                       = 27 8 (Shown) (iii) Req Prob  1 2 1 2 1 2P( 2, 2) 2 P( 2, 1) P( 2, 3)X X X X X X= = = + = = + = = 2 1 1 1 1 128 8 16 8 8        = + +               1 16=

(vi) The random variable Q is the product of the two numbers obtained from the two throws. Find the probability that Q is even, leaving your answer in terms of a. [2] (i) 0.4ab+= , since total probability =1 (ii) E( ) 3.5 3.5 0.1 2 0.6 0.8 5 0.6 1.4 2 5 X ab ab = = + + +