RI S2A Discrete Random Variables Tutorial Solns

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 __________________________________ Tutorial S2A: Discrete Random Variables Page 1 of 15 Tutorial S2A: Discrete Random Variables Section A (Discussion Questions) 1 Find the mean and standard deviation of the random variable X with the following probability distribution: x 4 3 2 1 0 1 2 3 P X x 0.04 0.16 0.24 0.16 0.15 0.1 0.1 0.05 [ –0.83, 1.86] E( ) P( )X x X x  =                  4 0.04 3 0.16 2 0.24 1 0.16 0 1 0.1 2 0.1 3 0.05           = –0.83 2 2E( ) P( )X x X x  = 4.15 2 2 Var( ) E( ) E( )X X X  =  24.15 0.83  = 3.4611   Standard deviation = 1.86 3 s.f.  Using GC, Enter data into 2 lists Read off the values from “sample mean”, x and “sample standard deviation”, x . E(X) = -0.83 SD(X) = 1.86 (3sf) Estimated number of tutorials: 2

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ __________________________________ Tutorial S2A: Discrete Random Variables Page 2 of 15 2 9233/1980/02/Q11 (modified) Two dice are thrown and the numbers A and B shown on each die are noted. The score X from the throw is defined by: if , if . A B A BX A B A B     (i) Tabulate the probability distribution of X. (ii) Evaluate the exact value of E(X) and Var(X). [(i) 28 9 (ii) 1015 162] 1 2 3 4 5 6 1 2 1 2 3 4 5 2 1 4 1 2 3 4 3 2 1 6 1 2 3 4 3 2 1 8 1 2 5 4 3 2 1 10 1 6 5 4 3 2 1 12 Probability distribution of X: x 1 2 3 4 5 6 8 10 12 P X x 10 5 36 18 9 1 36 4 6 1 36 6 5 36 2 1 36 18 1 36 1 36 1 36 1 36 E( ) P( )X x X x            10 9 6 5 2 1 11 2 3 4 5 6 836 36 36 36 36 36 36 1 1 10 1236 36                                                          = 28 9 2 2E( ) P( )X x X x            2 2 2 2 2 2 2 2 2 10 9 6 5 2 1 11 2 3 4 5 6 836 36 36 36 36 36 36 1 1 10 1236 36 287 18                                                            d1 d2 x

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ __________________________________ Tutorial S2A: Discrete Random Variables Page 3 of 15 2 2V ar( ) E( ) E( )X X X  2287 28 18 9       = 1015 162 The following GC screenshots are for checking purpose only (note that question asks for exact values of E(X) and Var(X)) 3 MJC Prelim 9233/2005/02/Q12 A discrete random variable X takes values 2, 3, 4, 5 with probabilities as shown in the table. x 2 3 4 5 P( )X x k 4 k 9 k 16 k (i) Find k, leaving your answer as a fraction. [2] (ii) Find E( cos )X , giving your answer to 3 significant figures. [3] (iii) Find 1 2P( 7)X X  , where 1X and 2X are two independent observations of X. [3] [(i) 144 205k (ii) 0.530 (iii) 0.0891] (i) 14 9 16 k k kk    144 205k  (ii)  cos 2 144 cos 3 36 cos 4 16 cos 5 9E cos 205X        0.530 (3 s.f.) (iii) 1 2 1 2 1 2P ( 7) 2P ( 2, 5) 2P ( 3, 4)X X X X X X        144 9 36 162 2205 205 205 205              3744 0.0891 (3 s.f.)42025 

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ __________________________________ Tutorial S2A: Discrete Random Variables Page 4 of 15 4 CJC Prelim 9233/2005/04/Q27 (modified) A box contains 2 fair tetrahedral dice. The first die has sides labeled 1, 1, 2, and 3 and the second die has sides labeled 1, 2, 3 and 3. A die is taken at random from the box and thrown. X, the score is defined as follows: If the first die is picked, then it is thrown and the score is defined as two times the number which appears on the base of the first die. If the second die is picked, then it is thrown and the score is the number which appears on the base of the second die. Show that P(X = 2) = 3 8 , and find the probability distribution of X. (i) Show that E(X) = 23 8 and find the exact value of Var(X). (ii) If this experiment is performed twice, find the probability that the score is 3 for the first experiment given that the total score is 4. (iii) Tom and Jerry take turns to perform this experiment with Jerry playing first. They stop when one of them obtains a score of 2. What is the probability of Tom obtaining a score of 2? [(i) 135 64 (ii) 2 13 (iii) 5 13 ] P(X = 2) = P(1st die picked & ‘1’ obtained) + P(2nd die picked & ‘2’ obtained) = 1 2  2 4 + 1 2  1 4 = 3 8 (shown) P(X = 1) = P(2nd die picked & ‘1’ obtained)= 1 2 1 4= 1 8 P(X = 3) = P (2nd die picked & ‘3’ obtained)= 1 2 2 4 = 1 4 P(X = 4) = P(1st die picked & ‘2’ obtained)= 2 4 1 4 = 1 8 P(X = 6) = P(1st die picked & ‘3’ obtained)= = 1 2  1 4 = 1 8 The probability distribution of X is: x 1 2 3 4 6 P(X = x) 1 8 3 8 1 4 1 8 1 8

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ __________________________________ Tutorial S2A: Discrete Random Variables Page 5 of 15 (i) E( X) = 1  1 8 + 2  3 8 + 3  1 4 + 4  1 8 + 6  1 8 = 23 8 E(X2) = 12 1 8 + 22 3 8 + 32 1 4 + 42 1 8 + 62 1 8 = 83 8 Var(X) = E(X2) – [ E(X) ]2 = 83 8 – ( 23 8 )2 = 135 64 (ii) P( X1 = 3 | X1 + X2 = 4) = P(X1 = 3 and X2 = 1) P(X1 + X2 = 4) = P(X1 = 3 and X2 = 1) 2 P(X1=1 & X2=3) + P(X1=2 & X2=2) =   1 4  1 8 2  1 8  1 4 +  3 8  3 8 = 2 13 (iii) Probability that Tom obtains a score of 2 3 55 3 5 3 5 3 8 8 8 8 8 8 15 15 56425 39 131 64                        

Raffles Institution H2 Mathematics 2025 Year 6 _____________________________________________________________________________________________ __________________________________ Tutorial S2A: Discrete Random Variables Page 6 of 15 5 9233/1993June/02/Q7(modified) Alfred and Bertie play a game, each starting with cash amounting to £100. Two dice are thrown. If the total score is 5 or more then Alfred pays £x, where 0 8x  , to Bertie. If the total score is 4 or less, then Bertie pays £(x + 8) to Alfred. By finding the probability distribution of Y, where Y denotes the random variable representing Alfred’s gains after one game, show that the expectation of Alfred’s cash after the first game is £1 3 (304 – 2x). Find the expectation of Alfred’s cash after six games. Find the value of x for the game to be fair. [Hint: For the game to be fair, expectation of Alfred’s gains needs to be equal to 0.] Given x = 3, find the variance of Alfred’s cash after the first game. [£(108 – 4x), x = 2, £2245 9 ] The probability distribution of Y is: y – x x + 8 P(Y = y) P score 5 5 6  6 1P score 436 6   Expectation of Alfred’s cash after one game = £E 100Y = £100 + EY = £  5 1100 + 86 6x x     