1. **State the problem:** We have a game where a player rolls a fair six-sided die. If the die shows 6, the player wins 40. If the die shows any other number (1 to 5), the player wins 0. The player pays 8 to play the game. We want to find the expected value of the game. 2. **Formula for expected value:** The expected value $E$ is calculated as $$E = \sum (\text{probability of outcome} \times \text{net winnings for that outcome})$$ 3. **Calculate probabilities and net winnings:** - Probability of rolling a 6: $\frac{1}{6}$ - Probability of rolling any other number: $\frac{5}{6}$ Net winnings are winnings minus cost to play: - If 6: $40 - 8 = 32$ - Otherwise: $0 - 8 = -8$ 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. **Cancel common factors:** $$E = \frac{\cancel{-8}}{\cancel{6}} = -\frac{4}{3}$$ 7. **Convert to decimal:** $$E = -1.33$$ **Final answer:** The expected value of the game is **-1.33** (rounded to the nearest cent). This means on average, the player loses 1.33 per game played.