Question

g.pr.10.6 (mc)

at a casino, there is a wheel with the numbers 1-50. the game costs \\$2 per spin. if you spin and land on the numbers 1-3, you win \\$1. if you spin the number 5 or 6, you win \\$10, and if you spin and land on 50, you win \\$25. the probability of winning each prize is below. find the probability of losing the game (spinning 7-49).

number spin and probability

\

$$\begin{tabular}{|c|c|c|c|c|} \\hline number spun & 1-3 & 5-6 & 50 & 7-49 \\\\ \\hline probability & 0.06 & 0.04 & 0.02 & \\\\ \\hline \\end{tabular}$$

(a) 0.12
(b) 0.88
(c) 0.5
(d) 1

Explanation:

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<reasoning>

Sum the probabilities of winning outcomes

\[
P(\text{winning}) = 0.06 + 0.04 + 0.02 = 0.12
\]

Calculate the probability of losing outcomes

\[
P(\text{losing}) = 1 - P(\text{winning}) = 1 - 0.12 = 0.88
\]
</reasoning>

<answer>
<mcq-option>(A) 0.12</mcq-option>
<mcq-correct>(B) 0.88</mcq-correct>
<mcq-option>(C) 0.5</mcq-option>
<mcq-option>(D) 1</mcq-option>
</answer>

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Answer:

<pre_analysis>
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</pre_analysis>

<reasoning>

Sum the probabilities of winning outcomes

\[
P(\text{winning}) = 0.06 + 0.04 + 0.02 = 0.12
\]

Calculate the probability of losing outcomes

\[
P(\text{losing}) = 1 - P(\text{winning}) = 1 - 0.12 = 0.88
\]
</reasoning>

<answer>
<mcq-option>(A) 0.12</mcq-option>
<mcq-correct>(B) 0.88</mcq-correct>
<mcq-option>(C) 0.5</mcq-option>
<mcq-option>(D) 1</mcq-option>
</answer>

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