You are given a bag containing six black coins and three whi

You are given a bag containing six black coins and three white coins and you are offered the following choice:

a) Draw one coin. If it is white, you win $5; if it is black you lose $1.

b) Draw two coins. If both are white, you win $10; if they are of different color you lose $2.

Assuming that you want to lay one of the two games, which game should you play? Show your work.

6 black coins an 3 white coins

a.
probability of drawing white coin = 3/(6+3) = 1/3
probability of drawing black coin = 2/3

expected earning = (1/3)*(5) + (2/3)*(-1)
= $ 1

b.
probability of drawing 2 white coins
= (3/9)*(2/8)
= 0.0833

probability that 2 coins are of different color
= P(first white second black OR first black second white)
= (3/9)*(6/8) + (6/9)*(3/8)
= 0.5

expected earning = 0.0833*10 + 0.5*(-2)
= -0.167

He should play game a)