Question

Suppose 2 dice are rolled, and suppose the discrete variable x counts the number of 3’s thats how up. Plot the probability distribution for x using a bar graph. What is the expected value for the random variable x? What is the variance for the random variable x?

Answer 1

Total number of possible outcomes when two dice are rolled is = 6² = 36.

The sample space is -

S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

If 'X' is the random variable representing the number of 3's that shows up then possible values of X are 0,1 and 2. Because we can either have no 3's, only one '3' or both the dice as '3'.

The possible cases are -

For no 3's = 5 x 5 (as both the dice can be any of the 5 values except 6) = 25 possible cases {(1,1), (1,2), (1,4), (1,5), (1,6), (2,1), (2,2), (2,4), (2,5), (2,6), (4,1), (4,2), (4,4), (4,5), (4,6), (5,1), (5,2), (5,4), (5,5), (5,6), (6,1), (6,2), (6,4), (6,5), (6,6)}

For one '3' = 2 x 5 (as any one of the dice has to be '3' and the other one strictly can't be '3') = 10 possible cases {(3,1), (3,2), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3), (6,3)}

And, for two 3's = 1 possible case (3,3)

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So, probability distribution of 'X' can be written as -

P(x=0) - 85 30 P(x = 1 ) : 1о 5 P(x: 2) = т з 36 18 25 36 5 18 for x=0 4 х= ) х=2 1 The box tab, л shown -

p(x) 1 2. O Х

Then E[] : 소깔 PX스z (3) + ()+ 2 () ㅋ I8 ] : 도개를. Px스가 (c ( ) (1) () + (2) 2> ( || + 11 5 18 t 니 18 9 에 18 2. Thus, Variance E[]- EL) 11 구요 8 에어 H9 324 3 32 2.