Question

3 white and 6 black balls in a box. Winner takes two balls at random. He gets 15 for each white ball and loses 5 for each black ball,. Find his expected gain

Answer 1

There are $3+6=9$ balls in a box. Winner takes two balls at random without replacement.

$$WW, WB, BW, BB$$

Let $X=$ gain.

$$P(WW)={3 \over 9}\cdot{2 \over 8}={1 \over 12},\ \ \ X=15+15=30$$

$$P(WB)={3 \over 9}\cdot{6 \over 8}={1 \over 4},\ \ \ X=15-5=10$$

$$P(BW)={6 \over 9}\cdot{3 \over 8}={1 \over 4},\ \ \ X=-5+15=10$$

$$P(BB)={6 \over 9}\cdot{5 \over 8}={5 \over 12},\ \ \ X=-5-5=-10$$

$${1 \over 12}+{1 \over 4}+{1 \over 4}+{5 \over 12}=1$$

Find his expected gain

$${1 \over 12}(30)+{1 \over 4}(10)+{1 \over 4}(10)+{5 \over 12}(-10)={10 \over 3}$$