1. **State the problem:** We have a game where a player rolls a die. If the die shows 6, the player wins 40. Otherwise, the player wins 0. The cost to play is 8. We want to find the expected value of the game and interpret it. 2. **Formula for expected value:** The expected value $E$ is calculated as: $$E = \sum (\text{probability of outcome} \times \text{net gain for outcome})$$ 3. **Calculate probabilities and net gains:** - Probability of rolling a 6: $\frac{1}{6}$ - Probability of not rolling a 6: $\frac{5}{6}$ - Net gain if 6 is rolled: $40 - 8 = 32$ - Net gain if not 6: $0 - 8 = -8$ 4. **Calculate expected value:** $$E = \frac{1}{6} \times 32 + \frac{5}{6} \times (-8)$$ 5. **Simplify:** $$E = \frac{32}{6} - \frac{40}{6} = \frac{32 - 40}{6} = \frac{-8}{6}$$ 6. **Simplify fraction with cancellation:** $$E = \frac{\cancel{-8}}{\cancel{6}} = -\frac{4}{3} = -1.33$$ 7. **Interpretation:** The expected value of $-1.33$ means the player expects to lose 1.33 per game on average over the long run. **Final answers:** - (a) The expected value is $-1.33$. - (b) The correct interpretation is option B: This value represents the expected loss over the long run for each game played.