Question

Let X equal the outcome (1, 2 , 3 or 4) when a fair four-sided die is rolled; let Y equal the outcome (1, 2, 3, 4, 5 or 6) when a fair six-sided die is rolled. Let W=X+Y.

a. What is the pdf of W?

b What is E(W)?

Answer 1

W = X + Y, and X + Y ranges from a minimum value of 2 (1 + 1) to a maximum value of 10 (4 + 6)

Total outcomes = 4 * 6 = 24

P(Sum = 2) = (1,1) = 1/24

P(Sum = 3) = (1,2) (2,1) = 2/24 = 1/12

P(Sum = 4) = (1,3) (3,1) (2,2) = 3/24 = 1/8

P(Sum = 5) = (1,4) (4,1) (2,3) (3,2) = 4/24 = 1/6

P(Sum = 6) = (1,5) (2,4) (4,2) (3,3) = 4/24 = 1/6

P(Sum = 7) = (1,6) (2,5) (3,4) (4,3) = 4/24 = 1/6

P(Sum = 8) = (2,6) (3,5) (4,4) = 3/24 = 1/8

P(Sum = 9) = (3,6) (4,5) = 2/24 = 1/12

P(Sum = 10) = (4,6) = 1/24

(a) Therefore the pdf is as below

W	2	3	4	5	6	7	8	9	10
P(W)	1/24	1/12	1/8	1/6	1/6	1/6	1/8	1/12	1/24

(b) Expected value = SUM [E - P(W) ] = (2 * 1/24) + (3 * 1/12) + (4 * 1/8) + (5 * 1/6) + (6 * 1/6) + 7 * (1/6) + (8 * 1/8) + (9 * 1/12) + (10 * 1/24)

Converting everything to a common denominator of 24

= 2/24 + 6/24 + 12/24 + 20/24 + 24/24 + 28/24 + 24/24 + 18/24 + 10/24 = 144/24 = 6