Question

g.pr.10.6 (mc)

florence wants to play a game at a casino where she will roll a fair die and spin a spinner with the colors red, blue, green, and yellow evenly spread out. the game costs \\$5 to play and there are 3 ways to win.

• roll a 1 or 2 and spin red or yellow-win \\$2
• roll a 3 and spin blue or red-win \\$20
• roll an even number and spin a yellow-win \\$5

find the probability of not winning a prize.

\\(\bigcirc\\) \\(\frac{1}{3}\\)
\\(\bigcirc\\) \\(\frac{5}{8}\\)
\\(\bigcirc\\) \\(\frac{2}{3}\\)
\\(\bigcirc\\) \\(\frac{3}{4}\\)

Explanation:

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<reasoning>

Calculate the probability of each winning event

\[

$$\begin{aligned} P(\text{Win 1}) &= P(\text{Roll } 1 \text{ or } 2) \times P(\text{Spin red or yellow}) = \frac{2}{6} \times \frac{2}{4} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \\ P(\text{Win 2}) &= P(\text{Roll } 3) \times P(\text{Spin blue or red}) = \frac{1}{6} \times \frac{2}{4} = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \\ P(\text{Win 3}) &= P(\text{Roll even}) \times P(\text{Spin yellow}) = \frac{3}{6} \times \frac{1}{4} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \end{aligned}$$

\]

Sum the probabilities of winning any prize

\[

$$\begin{aligned} P(\text{Win}) &= P(\text{Win 1}) + P(\text{Win 2}) + P(\text{Win 3}) \\ &= \frac{1}{6} + \frac{1}{12} + \frac{1}{8} \\ &= \frac{4}{24} + \frac{2}{24} + \frac{3}{24} = \frac{9}{24} = \frac{3}{8} \end{aligned}$$

\]

Calculate the probability of not winning a prize

\[
P(\text{Not Win}) = 1 - P(\text{Win}) = 1 - \frac{3}{8} = \frac{5}{8}
\]
</reasoning>

<answer>
<mcq-option>(A) \(\frac{1}{3}\)</mcq-option>
<mcq-correct>(B) \(\frac{5}{8}\)</mcq-correct>
<mcq-option>(C) \(\frac{2}{3}\)</mcq-option>
<mcq-option>(D) \(\frac{3}{4}\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Calculate the probability of each winning event

\[

$$\begin{aligned} P(\text{Win 1}) &= P(\text{Roll } 1 \text{ or } 2) \times P(\text{Spin red or yellow}) = \frac{2}{6} \times \frac{2}{4} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \\ P(\text{Win 2}) &= P(\text{Roll } 3) \times P(\text{Spin blue or red}) = \frac{1}{6} \times \frac{2}{4} = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} \\ P(\text{Win 3}) &= P(\text{Roll even}) \times P(\text{Spin yellow}) = \frac{3}{6} \times \frac{1}{4} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \end{aligned}$$

\]

Sum the probabilities of winning any prize

\[

$$\begin{aligned} P(\text{Win}) &= P(\text{Win 1}) + P(\text{Win 2}) + P(\text{Win 3}) \\ &= \frac{1}{6} + \frac{1}{12} + \frac{1}{8} \\ &= \frac{4}{24} + \frac{2}{24} + \frac{3}{24} = \frac{9}{24} = \frac{3}{8} \end{aligned}$$

\]

Calculate the probability of not winning a prize

\[
P(\text{Not Win}) = 1 - P(\text{Win}) = 1 - \frac{3}{8} = \frac{5}{8}
\]
</reasoning>

<answer>
<mcq-option>(A) \(\frac{1}{3}\)</mcq-option>
<mcq-correct>(B) \(\frac{5}{8}\)</mcq-correct>
<mcq-option>(C) \(\frac{2}{3}\)</mcq-option>
<mcq-option>(D) \(\frac{3}{4}\)</mcq-option>
</answer>

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