Question

Bin $A$ has one white ball and four black balls. Bin $B$ has three balls labeled $\$1$ and one ball labeled $\$7$. Bin $W$ has five balls labeled $\$8$ and one ball labeled $\$500$. A game is played as follows: a ball is randomly selected from bin $A$. If it is black, then a ball is randomly selected from bin $B$; otherwise, if the original ball is white, then a ball is randomly selected from bin $W$. You win the amount printed on the second ball selected. What is your expected win

Answer 1

$20

Explanation:

Since Bin A has one white ball and four black balls, the money ball has a 1/5 chance of coming from Bin W and a 4/5 chance of coming from Bin B. The total expected value therefore is $E = 1/5E_W+4/5E_B, where E_W and E_B are the expected values of a ball drawn from bins W and B, respectively. Since Bin W has five 8 dollar balls and one 500 dollar ball, its expected value is E_W = 5/6*8 + 1/6*500 = 90. Since Bin B has three 1 dollar balls and one 7 dollar ball, its expected value is E_B = 3/4*1 + 1/4*7 = $2.5. Therefore E = 1/5E_W + 4/5E_B = 1/5*90 + 4/5*2.5 = 20

Answer 2

$20

Explanation:

Bin A: Bin B: Bin W:

white ball: 20% $1: 75% $8: 5/6

black ball: 80% $7: 25% $500: 1/6

first we can calculate the expected return of bins B and W:

expected return if the ball is black = ($1 x 75%) + ($7 x 25%) = $2.50

expected return if the ball is white = ($8 x 5/6) + ($500 x 1/6) = $90

expected return of the game = (expected return of a black ball x probability of choosing a black ball) + (expected return of a white ball x probability of choosing a white ball) =($2.50 x 80%) + ($90 x 20%) = $2 + $18 = $20