A game is played using one die. If the die is rolled and shows 6 , the player wins \( \$ 35 \). If the dife shows any number other than 6 , the player wins nothing. Complete parts (a) through (b) below. a. If there is a charge of \( \$ 7 \) to play the game, what is the game's expected value? \( \$-1.17 \) (Round to the nearest cent.) b. What does this expected value mean? Choose the correct statement below. A. The player can expect to break even over the long run. B. This value represents the expected win over the long run for each game played. c. This value represents the expected loss over the long run for each game played.

To determine the expected value of the game, we first need to understand the possible outcomes. The probability of rolling a 6 is \( \frac{1}{6} \), leading to a win of \( \$35 \), and the probability of rolling any other number (1-5) is \( \frac{5}{6} \), resulting in a win of \( \$0 \). After factoring in the cost to play, the expected value is calculated as follows: \[ \text{Expected Value} = \left(\frac{1}{6} \times 35\right) + \left(\frac{5}{6} \times 0\right) - 7 = 5.83 - 7 = -1.17 \] This means that, on average, players lose about \( \$1.17 \) each time they play. The expected value indicates that, over the long term, the game is not favorable to the player. Thus, the correct statement is C. This value represents the expected loss over the long run for each game played. Keep this in mind: while the thrill of winning can be tempting, the house always has an edge!