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.
