A game consists of tossing a coin and rolling an ordinary die. If the player tosses a head and rolls an even number, the player wins $3. Otherwise, the player loses $1. What is the expected value of this game?

Select one:
a. $6
b. $0
c. $3
d. $1

$0

To find the expected value of a game, we need to calculate the value for each possible outcome and then multiply it by its corresponding probability.

In this game, there are 6 possible outcomes:
1. Tails and an odd number on the die (no win, -$1)
2. Tails and an even number on the die (no win, -$1)
3. Heads and an odd number on the die (no win, -$1)
4. Heads and an even number on the die (win, +$3)
5. Tails and a 1 on the die (no win, -$1)
6. Tails and a 6 on the die (no win, -$1)

To calculate the expected value, we need to assign probabilities to each outcome. Since the coin and the die are both fair and independent, each outcome has a probability of 1/2 * 1/6 = 1/12.

Now, let's calculate the expected value:

Expected value = (1/12 * -$1) + (1/12 * -$1) + (1/12 * -$1) + (1/12 * $3) + (1/12 * -$1) + (1/12 * -$1)
= (-$1/12) + (-$1/12) + (-$1/12) + ($3/12) + (-$1/12) + (-$1/12)
= -$6/12 + $3/12 - $6/12
= -$9/12
= -$0.75

Therefore, the expected value of this game is -$0.75, which means the player can expect to lose an average of $0.75 per game.

So the correct answer is b. $0