Question

In a particular game, a fair die is tossed. If the number of spots showing is a six, you win $6, if the number of spots showing is a five, you win $3, if the number of spots showing is 4, you win $2, and if the number of spots showing is 3, you win $1. If the number of spots showing is 1 or 2, you win nothing. You are going to play the game twice. Let X be the money you win in the first game, and let y be the money you win in the second game. Find Var[X+Y] A) 4.00 B) 4.33. C) 6.00 D) 8.67.

Answer 1

Outcome on dice	6	5	4	3	1 or 2
Amount (x)	6	3	2	1	0
Probability (p)	1/6	1/6	1/6	1/6	1/3

$\tiny Var(X) = E(x^2) - (E(x))^2 = 6^2 *\frac{1}{6}+3^2 *\frac{1}{6}+2^2*\frac{1}{6}+1^2 *\frac{1}{6}+0 - (6 *\frac{1}{6}+3 *\frac{1}{6}+2*\frac{1}{6}+1 *\frac{1}{6}+0*\frac{1}{6})^2$$\tiny Var(X) = \frac{13}{3}$

As X and Y both are same thing so Var(x) = Var(y) = 13/3

Since both X and Y are independent of each other.

Hence, Var(X + Y) = Var(X) + Var(Y) = (13/3) + (13/3) = 8.67 Option D)

Please comment if any doubt. Thank you.