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.

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UpStudy AI · Accepted Answer

To solve the problem, we need to calculate the expected value of the game, taking into account the charge to play and the potential winnings.

### Part (a): Calculate the Expected Value

1. **Identify the outcomes and their probabilities:**
   - The die has 6 faces, so the probability of rolling a 6 (winning) is $$ P(6) = \frac{1}{6} $$.
   - The probability of rolling any number other than 6 (losing) is $$ P(\text{not } 6) = \frac{5}{6} $$.

2. **Determine the winnings and costs:**
   - If the player rolls a 6, they win $$ \$35 $$, but since there is a charge of $$ \$7 $$ to play, the net gain is $$ 35 - 7 = 28 $$.
   - If the player rolls any other number, they win nothing, so the net loss is $$ -7 $$.

3. **Calculate the expected value (EV):**
   $$
   EV = (P(6) \times \text{Net Gain}) + (P(\text{not } 6) \times \text{Net Loss})
   $$
   Substituting the values:
   $$
   EV = \left(\frac{1}{6} \times 28\right) + \left(\frac{5}{6} \times (-7)\right)
   $$

4. **Perform the calculations:**
   - Calculate the first term: 
   $$
   \frac{1}{6} \times 28 = \frac{28}{6} \approx 4.67
   $$
   - Calculate the second term:
   $$
   \frac{5}{6} \times (-7) = -\frac{35}{6} \approx -5.83
   $$
   - Combine the two results:
   $$
   EV \approx 4.67 - 5.83 \approx -1.17
   $$

Thus, the expected value of the game is approximately $$ \$-1.17 $$.

### Part (b): Interpret the Expected Value

The expected value of $$ \$-1.17 $$ means that, on average, a player can expect to lose $$ \$1.17 $$ for each game played in the long run.

The correct statement is:
**C. This value represents the expected loss over the long run for each game played.**
The game's expected value is $$ \$-1.17 $$, meaning the player can expect to lose \$1.17 on average per game over the long run.

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