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Sagot :
Sure, let's analyze the game step-by-step to find the expected value.
First, let’s identify the outcomes and their respective probabilities:
1. Outcome 1: Roll a 1, win [tex]$2.00 2. Outcome 2, 3, 4: Roll a 2, 3, or 4, lose $[/tex]1.00 each time
3. Outcome 5, 6: Roll a 5 or 6, win [tex]$1.00 Now, calculate the probabilities for rolling each possible outcome: 1. Probability of rolling a 1: Since there is only one favorable outcome and a die has 6 faces, the probability is \( \frac{1}{6} \approx 0.167 \). 2. Probability of rolling a 2, 3, or 4: There are three favorable outcomes, so the probability is \( \frac{3}{6} = 0.500 \). 3. Probability of rolling a 5 or 6: There are two favorable outcomes, so the probability is \( \frac{2}{6} \approx 0.333 \). Next, we need to multiply each outcome by its respective probability to find the products: 1. Product for rolling a 1: \( 2.00 \times 0.167 = 0.33 \) 2. Product for rolling a 2, 3, or 4: \( -1.00 \times 0.500 = -0.50 \) 3. Product for rolling a 5 or 6: \( 1.00 \times 0.333 = 0.33 \) Lastly, we'll sum these products to find the expected value: Expected value = \( 0.33 - 0.50 + 0.33 = 0.16 \) Now, let's complete the table with these calculated values: \[ \begin{tabular}{|c|c|c|c|} \hline Result & Outcome $[/tex](X)[tex]$ & Probability $[/tex]P(X)[tex]$ & Product $[/tex]X \cdot P(X)[tex]$ \\ \hline 1 & \$[/tex]2.00 & 0.167 & \[tex]$0.33 \\ \hline 2,3,4 & -\$[/tex]1.00 & 0.500 & -\[tex]$0.50 \\ \hline 5,6 & \$[/tex]1.00 & 0.333 & \$0.33 \\
\hline
\end{tabular}
\]
Final expected value: [tex]\(0.16\)[/tex].
This table summarizes the expected value calculations for this carnival game.
First, let’s identify the outcomes and their respective probabilities:
1. Outcome 1: Roll a 1, win [tex]$2.00 2. Outcome 2, 3, 4: Roll a 2, 3, or 4, lose $[/tex]1.00 each time
3. Outcome 5, 6: Roll a 5 or 6, win [tex]$1.00 Now, calculate the probabilities for rolling each possible outcome: 1. Probability of rolling a 1: Since there is only one favorable outcome and a die has 6 faces, the probability is \( \frac{1}{6} \approx 0.167 \). 2. Probability of rolling a 2, 3, or 4: There are three favorable outcomes, so the probability is \( \frac{3}{6} = 0.500 \). 3. Probability of rolling a 5 or 6: There are two favorable outcomes, so the probability is \( \frac{2}{6} \approx 0.333 \). Next, we need to multiply each outcome by its respective probability to find the products: 1. Product for rolling a 1: \( 2.00 \times 0.167 = 0.33 \) 2. Product for rolling a 2, 3, or 4: \( -1.00 \times 0.500 = -0.50 \) 3. Product for rolling a 5 or 6: \( 1.00 \times 0.333 = 0.33 \) Lastly, we'll sum these products to find the expected value: Expected value = \( 0.33 - 0.50 + 0.33 = 0.16 \) Now, let's complete the table with these calculated values: \[ \begin{tabular}{|c|c|c|c|} \hline Result & Outcome $[/tex](X)[tex]$ & Probability $[/tex]P(X)[tex]$ & Product $[/tex]X \cdot P(X)[tex]$ \\ \hline 1 & \$[/tex]2.00 & 0.167 & \[tex]$0.33 \\ \hline 2,3,4 & -\$[/tex]1.00 & 0.500 & -\[tex]$0.50 \\ \hline 5,6 & \$[/tex]1.00 & 0.333 & \$0.33 \\
\hline
\end{tabular}
\]
Final expected value: [tex]\(0.16\)[/tex].
This table summarizes the expected value calculations for this carnival game.
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