Two balls are chosen randomly from an urn containing 8 yello

Two balls are chosen randomly from an urn containing 8 yellow, 4 black, and 2 red balls. Let Y represent the number of black balls chosen. The discrete probability distribution is shown below.

P(Y=0) = 45/91
P(Y=1) = 40/91
P(Y=2) = 6/91

What is the variance of Y?

With the situation described above, now suppose that you win $2 for each black ball selected and nothing for any other color. What is the variance of your winnings?

Solution

a.) E(Y) = Sum of y*P(y) = (0)*(45/91) + (1)*(40/91) + (2)*(6/91) = 4/7

E(Y^2) = Sum of y^2*P(y) = (0)*(45/91) + (1)*(40/91) + (4)*(6/91) = 64/91

Var(Y) = E(Y^2) - (E(Y))^2 = (64/91) - (4/7)^2 = 240/637

b.) Variance of winnings = ($2)^2*(240/637) = ($4)*(240/637) = $960/637 = $1.5071