At a county fair there is a spinner game with 10 sectors: 2 red sectors, 2 green sectors, 3 blue sectors, and 3 yellow sectors. If the spinner lands on a red sector, the player wins 3 tokens. If the spinner lands on a green sector, the player wins 5 tokens. If the spinner lands on any other sector, the player loses 2 tokens.
Is this game fair for the player and how much will the player win or lose on an average over time?
A)County fair games are never fair.
B)The expected value is zero, and the game is fair. So the player will break even for a single spin over time.
C)The expected value is 0.4 and the game is unfair. The player will win about 0.4 tokens for a single spin over time.
D)The expected value is -0.4 and the game is not fair. The player will lose about 0.4 tokens for a single spin over time.
+7 There is a \(\frac{2}{10}\) chance (probability of landing on a red sector) that the player wins 3 tokens, and \(\frac{2}{10} \cdot 3 = \frac{6}{10} = \frac35\).
There is a \(\frac{2}{10}\) chance (probability of landing on a green sector) that the player wins 5 tokens, and \(\frac{2}{10} \cdot 5 = \frac{10}{10} = 1\).
There is a is a \(\frac{6}{10}\) chance (probability of landing on a blue or yellow sector) that the player loses 2 tokens, and \(\frac{6}{10} \cdot (-2) = -\frac{12}{10} = -\frac{6}{5}\).
Adding these numbers up, we get \(\frac{3}{5} + 1 - \frac{6}{5} = \frac{2}{5}\), so the answer is \(\boxed{\text{C}}\).
+7 There is a \(\frac{2}{10}\) chance (probability of landing on a red sector) that the player wins 3 tokens, and \(\frac{2}{10} \cdot 3 = \frac{6}{10} = \frac35\).
There is a \(\frac{2}{10}\) chance (probability of landing on a green sector) that the player wins 5 tokens, and \(\frac{2}{10} \cdot 5 = \frac{10}{10} = 1\).
There is a is a \(\frac{6}{10}\) chance (probability of landing on a blue or yellow sector) that the player loses 2 tokens, and \(\frac{6}{10} \cdot (-2) = -\frac{12}{10} = -\frac{6}{5}\).
Adding these numbers up, we get \(\frac{3}{5} + 1 - \frac{6}{5} = \frac{2}{5}\), so the answer is \(\boxed{\text{C}}\).
