Question

A local high school runs the following game at an

annual fundraising event. In this game, marbles
are randomly picked from a bag. The bag contains
four red marbles, three white marbles, and one
black marble. If you pick the black marble,
you win $5. If the one you pick is white, you
win $2. If you pick a red one, you do not win
anything. The game costs $1.50 to play. What is
the expected value of the game?
A. -$0.125
C. $0.50
B. $0.375
D. -$0.50

Answer 1

The expected value of the game can be calculated by multiplying the probability of each outcome by its respective payoff, and then adding these products together.

The probability of picking the black marble is 1/8, and the payoff is $5. Therefore, the expected value of picking the black marble is (1/8) x $5 = $0.625.

The probability of picking a white marble is 3/8, and the payoff is $2. Therefore, the expected value of picking a white marble is (3/8) x $2 = $0.75.

The probability of picking a red marble is 4/8, and the payoff is -$1.50 (since the game costs $1.50 to play). Therefore, the expected value of picking a red marble is (4/8) x (-$1.50) = -$0.75.

Finally, we add the expected values of each outcome together: $0.625 + $0.75 - $0.75 = $0.625.

Therefore, the expected value of the game is $0.625. Answer: C. $0.50.