H2MA Remedial DRV Solution (RVHS)

Teacher’s copy 1 RVHS H2 Mathematics Remedial Programme Topic: Discrete Random Variables Basic Mastery Questions 1. CJC MYE 9758/2021//Q9 (Parts) A biased red die is such that the probability of any face landing up wards is proportional to the square of the number on that face. The random variable X denotes the score obtained in one throw of this die with ( ) 2P X r kr== , where 1r = , 2, 3, 4, 5, 6, and k is a constant. (i) Find the exact value of k . [2] A second biased die is yellow and the random variable Y denotes the score obtained when the yellow die is thrown once. The probability distribution of Y is y 2 4 6 ( )P Yy= 1 5 2 5 2 5 (ii) Find ( )E Y and show that ( ) 56Var 25Y = . [3] (iii) Given that 1Y and 2Y are two independent observations of Y , find ( )12E YY− and ( )12Var YY− . [2] Answer: (i) 1 91k = (ii) 22 5 (iii) 0, 4.48 (i) r 1 2 3 4 5 6 ( )P Xr= k 4k 9k 16k 25k 36k 4 9 16 25 36 1 91 1 1 91 k k k k k k k k + + + + + = = = (ii) ( ) 1 2 2 22E 2 4 6 5 5 5 5Y      = + + =           ( ) 2 2 2 2 1 2 2 108E 2 4 6 5 5 5 5Y      = + + =          

Teacher’s copy 2 ( ) ( ) ( ) ( ) 22 2 Var =E E 108 22 55 56 shown25 Y Y Y −  =−  = (iii) ( ) ( ) ( )1 2 1 2 22 22E E E 0 55Y Y Y Y− = − = − = ( ) ( ) ( ) ( )1 2 1 2 56 56 112Var Var Var or 4.48 25 25 25Y Y Y Y− = + = + = 2. RI MYE 9758/2021//Q10(i) In a game, a player tosses a fair die, whose faces are numbered from 1 to 6. If the player obtains a 6, he tosses the die a second time, and in this case, his score is the absolute difference of 6 and the second number. Otherwise, his score is the number obtained in the first toss. Let the player’s score be denoted by X. Show that ( ) 7P1 36X == and tabulate the probability distribution of X. [3] ( )P 1 P(first throw = 1) + P(1st throw = 6, 2nd throw = 5) 1 1 1 = 6 6 6 11 6 36 7 (Shown)36 X ==   +     =+ = Probability distribution of X x 0 1 2 3 4 5 P( )Xx= 1 36 7 36 7 36 7 36 7 36 7 36

Teacher’s copy 3 Standard Questions 1. HCI MYE 9758/2020//Q8 (Parts) A bag contains 9 numbered balls of identical size. Four of the balls are numbered 3, three of the balls are numbered 4 and two of the balls are numbered 5. In a game, three balls are drawn from the bag at random, without replacement. The random variable S is the sum of the numbers on the three balls drawn. (i) Show that ( ) 25P 12 84S == and find the probability distribution of S. [4] (ii) Show that the probability where the sum of the numbers on the three balls drawn is a multiple of 3 is given by 29 84 . [1] (i) When listing out all the outcomes, do it systematically: (a) All 3 balls have the same number 3+3+3 = 9 4+4+4 = 12 5+5+5 = 15 (b) Only 2 balls have the same number 3+3+(4 or 5) = 10 or 11 4+4+(3 or 5) = 11 or 13 5+5+(3 or 4) = 13 or 14 (c) All 3 balls different numbers 3+4+5 = 12 ( ) 34 3 2 31 1 1 99 33 25P 12 84 CC C CS CC = = + = Or ( ) 4 3 2 3 2 1 25P 12 3! 9 8 7 9 8 7 84S    = =    +   =       (shown) ( ) 4 3 9 3 1P9 21 CS C= = = Or ( ) 4 3 2 1P9 9 8 7 21S = =   = ( ) 43 21 9 3 3P 10 14 CCS C = = = Or ( ) 4 3 3 3P 10 3 9 8 7 14S = =    =

Teacher’s copy 4 ( ) 4 2 4 3 2 1 1 2 99 33 2P 11 7 C C C CS CC = = + = Or ( ) 4 3 2 3 2 4 2P 11 3 3 9 8 7 9 8 7 7S    = =    +    =       ( ) 4 2 3 2 1 2 2 1 99 33 5P 13 42 C C C CS CC = = + = Or ( ) 3 2 2 2 1 4 5P 13 3 3 9 8 7 9 8 7 42S    = =    +    =       ( ) 32 12 9 3 1P 14 28 CCS C = = = Or ( ) 2 1 3 1P 14 3 9 8 7 28S = =    = s 9 10 11 12 13 14 ( )P Ss= 1 21 3 14 2 7 25 84 5 42 1 28 (ii) ( ) ( )Required probability P 9 P 12 1 25 29 21 84 84 SS= = + = = + =

Teacher’s copy 5 2. RVHS MYE 9758/2020//Q8 In a funfair game, a game-master set up two boxes with each box containing four cards, numbered 1, 2, 3, 4. A player draws one card at random from each box and his score X, is the product of the numbers on the two cards. (i) Find the probability distribution of X. [2] (ii) Calculate the mean score and the variance exactly. [2] The game-master charges $p for each game. If the player’s score is odd, the player wins a $5 cash voucher. Otherwise, the game ends. (iii) Find the range of values of p for the game to be in favour of the game-master. [2] Answer: (ii) 6.25, 17.1875 (iii) 5 3p (i) Probability distribution of X: x 1 2 3 4 6 8 9 12 16 P(X = x) 1 16 2 16 2 16 3 16 2 16 2 16 1 16 2 16 1 16 (ii) 1E( ) (1 1 2 2 3 2 4 3 6 216 +8 2 9 1 12 2 16 1) 6.25 X =  +  +  +  +   +  +  +  = 2 2 2 2 2 2 2 2 2 2 1E( ) (1 1 2 2 3 2 4 3 6 216 8 2 9 1 12 2 16 1) 56.25 X =  +  +  +  +  +  +  +  +  = Var(X) = E(X2) − E(X)2 = 17.1875 (iii) Let Y denote the gain of the game-master. For the game to be in favour of game-master, E(Y) > 0

Teacher’s copy 6 P( is even) ( 5) P( is odd) 0 31 5044 5 (or 1.67)3 p X X p pp   + −       −          