Question

g.pr.10.6 (hc)

a game is played where you draw a card from a standard deck and spin a wheel with the numbers 1-10. it costs $1 to play. the winning plays are below:

• draw a face card and spin a 2-win $5
• draw a queen and spin an odd number-win $10
• draw a red king and spin a 10-win $50

determine the expected value for the game.

( ) -$0.31
( ) $11.25
( ) -0.25
( ) $15

Explanation:

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<reasoning>

Calculate probabilities of winning events

\[

$$\begin{aligned} &P(\text{Face Card and } 2) = \frac{12}{52} \times \frac{1}{10} = \frac{3}{13} \times \frac{1}{10} = \frac{3}{130} \\ &P(\text{Queen and Odd}) = \frac{4}{52} \times \frac{5}{10} = \frac{1}{13} \times \frac{1}{2} = \frac{1}{26} = \frac{5}{130} \\ &P(\text{Red King and } 10) = \frac{2}{52} \times \frac{1}{10} = \frac{1}{26} \times \frac{1}{10} = \frac{1}{260} \end{aligned}$$

\]

Calculate expected winnings

\[

$$\begin{aligned} E(\text{Winnings}) &= 5 \times P(\text{Face Card and } 2) + 10 \times P(\text{Queen and Odd}) + 50 \times P(\text{Red King and } 10) \\ &= 5 \times \frac{3}{130} + 10 \times \frac{5}{130} + 50 \times \frac{1}{260} \\ &= \frac{15}{130} + \frac{50}{130} + \frac{25}{130} \\ &= \frac{90}{130} = \frac{9}{13} \approx 0.6923 \end{aligned}$$

\]

Calculate expected value of the game

\[

$$\begin{aligned} E(\text{Game}) &= E(\text{Winnings}) - \text{Cost to play} \\ &= 0.6923 - 1.00 \\ &\approx -0.3077 \approx -\$0.31 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) -\$0.31</mcq-correct>
<mcq-option>(B) \$11.25</mcq-option>
<mcq-option>(C) -0.25</mcq-option>
<mcq-option>(D) \$15</mcq-option>
</answer>

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Answer:

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<reasoning>

Calculate probabilities of winning events

\[

$$\begin{aligned} &P(\text{Face Card and } 2) = \frac{12}{52} \times \frac{1}{10} = \frac{3}{13} \times \frac{1}{10} = \frac{3}{130} \\ &P(\text{Queen and Odd}) = \frac{4}{52} \times \frac{5}{10} = \frac{1}{13} \times \frac{1}{2} = \frac{1}{26} = \frac{5}{130} \\ &P(\text{Red King and } 10) = \frac{2}{52} \times \frac{1}{10} = \frac{1}{26} \times \frac{1}{10} = \frac{1}{260} \end{aligned}$$

\]

Calculate expected winnings

\[

$$\begin{aligned} E(\text{Winnings}) &= 5 \times P(\text{Face Card and } 2) + 10 \times P(\text{Queen and Odd}) + 50 \times P(\text{Red King and } 10) \\ &= 5 \times \frac{3}{130} + 10 \times \frac{5}{130} + 50 \times \frac{1}{260} \\ &= \frac{15}{130} + \frac{50}{130} + \frac{25}{130} \\ &= \frac{90}{130} = \frac{9}{13} \approx 0.6923 \end{aligned}$$

\]

Calculate expected value of the game

\[

$$\begin{aligned} E(\text{Game}) &= E(\text{Winnings}) - \text{Cost to play} \\ &= 0.6923 - 1.00 \\ &\approx -0.3077 \approx -\$0.31 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) -\$0.31</mcq-correct>
<mcq-option>(B) \$11.25</mcq-option>
<mcq-option>(C) -0.25</mcq-option>
<mcq-option>(D) \$15</mcq-option>
</answer>

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