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Sagot :
Sure! Let's go through the problem step-by-step:
### Part (a):
We need to determine the possible outcomes and their probabilities.
#### Possible Outcomes
1. (1, 3)
2. (1, 5)
3. (1, 1)
4. (3, 5)
5. (3, 3)
6. (5, 5)
#### X (Amount Miguel wins or loses)
- Miguel wins [tex]$2 if the chips match: (1, 1), (3, 3), (5, 5) - Miguel loses $[/tex]1 if the chips do not match: (1, 3), (1, 5), (3, 5)
#### P(x) (Probability of each outcome)
- Probability of winning [tex]$2: - Outcomes: (1, 1), (3, 3), (5, 5) - Count: 1 (for each pairing) - Total: 3 possible outcomes - Probability of losing $[/tex]1:
- Outcomes: (1, 3), (1, 5), (3, 5)
- Count: 1 (for each pairing)
- Total: 3 possible outcomes
Since there are 6 possible outcomes in total, the probability for each scenario is:
- [tex]\( P(\text{win } \$2) = \frac{3}{6} = 0.3333 \)[/tex]
- [tex]\( P(\text{lose } -\$1) = \frac{3}{6} = 0.6667 \)[/tex]
\begin{tabular}{|c|c|c|}
\hline
[tex]$x$[/tex] & 2 & -1 \\
\hline
[tex]$P(x)$[/tex] & 0.3333 & 0.6667 \\
\hline
\end{tabular}
### Part (b):
Miguel's expected value from playing the game is calculated by:
[tex]\[ E(X) = \sum (x_i \cdot P(x_i)) \][/tex]
Given:
[tex]\[ x = 2 \quad \text{with } P(x) = 0.3333 \][/tex]
[tex]\[ x = -1 \quad \text{with } P(x) = 0.6667 \][/tex]
Then,
[tex]\[ E(X) = (2 \cdot 0.3333) + (-1 \cdot 0.6667) \][/tex]
[tex]\[ E(X) = 0.6666 - 0.6667 \][/tex]
[tex]\[ E(X) = 0.0 \][/tex]
### Part (c):
Based on the expected value calculated above, Miguel should expect to neither win nor lose any money on average with each game he plays. The expected value is [tex]$0, which means Miguel breaks even over time. ### Part (d): To make the game fair (expected value of 0), we need to assign a fair value to the winning amount when Miguel picks two chips with the number 1. We denote this amount as \( y \). The fair condition sets up the equation as: \[ P(\text{win } 2) \times 2 + P(\text{lose } -1) \times -1 + P(\text{win } y) \times y = 0 \] Since \( P(\text{win } y) \) is the probability of picking (1, 1), which is \(\frac{1}{6}\): \[ 0.3333 \times 2 + 0.6667 \times -1 + \frac{1}{6} \times y = 0 \] Solving for \( y \): \[ 0.6666 - 0.6667 + 0.1667y = 0 \] \[ 0.1667y = 0 - (0.6667 - 0.6666) \] \[ 0.1667y = 0.0001 \] \[ y = \frac{0.0001}{0.1667} \approx 0.0 \] Therefore, \( y = 0 \) is the value that would make the game fair. Miguel would neither be expected to win nor lose on average. In a complete sentence, "To make the game fair, the amount Miguel would win if he picks two chips with the number 1 should be $[/tex]0."
### Part (a):
We need to determine the possible outcomes and their probabilities.
#### Possible Outcomes
1. (1, 3)
2. (1, 5)
3. (1, 1)
4. (3, 5)
5. (3, 3)
6. (5, 5)
#### X (Amount Miguel wins or loses)
- Miguel wins [tex]$2 if the chips match: (1, 1), (3, 3), (5, 5) - Miguel loses $[/tex]1 if the chips do not match: (1, 3), (1, 5), (3, 5)
#### P(x) (Probability of each outcome)
- Probability of winning [tex]$2: - Outcomes: (1, 1), (3, 3), (5, 5) - Count: 1 (for each pairing) - Total: 3 possible outcomes - Probability of losing $[/tex]1:
- Outcomes: (1, 3), (1, 5), (3, 5)
- Count: 1 (for each pairing)
- Total: 3 possible outcomes
Since there are 6 possible outcomes in total, the probability for each scenario is:
- [tex]\( P(\text{win } \$2) = \frac{3}{6} = 0.3333 \)[/tex]
- [tex]\( P(\text{lose } -\$1) = \frac{3}{6} = 0.6667 \)[/tex]
\begin{tabular}{|c|c|c|}
\hline
[tex]$x$[/tex] & 2 & -1 \\
\hline
[tex]$P(x)$[/tex] & 0.3333 & 0.6667 \\
\hline
\end{tabular}
### Part (b):
Miguel's expected value from playing the game is calculated by:
[tex]\[ E(X) = \sum (x_i \cdot P(x_i)) \][/tex]
Given:
[tex]\[ x = 2 \quad \text{with } P(x) = 0.3333 \][/tex]
[tex]\[ x = -1 \quad \text{with } P(x) = 0.6667 \][/tex]
Then,
[tex]\[ E(X) = (2 \cdot 0.3333) + (-1 \cdot 0.6667) \][/tex]
[tex]\[ E(X) = 0.6666 - 0.6667 \][/tex]
[tex]\[ E(X) = 0.0 \][/tex]
### Part (c):
Based on the expected value calculated above, Miguel should expect to neither win nor lose any money on average with each game he plays. The expected value is [tex]$0, which means Miguel breaks even over time. ### Part (d): To make the game fair (expected value of 0), we need to assign a fair value to the winning amount when Miguel picks two chips with the number 1. We denote this amount as \( y \). The fair condition sets up the equation as: \[ P(\text{win } 2) \times 2 + P(\text{lose } -1) \times -1 + P(\text{win } y) \times y = 0 \] Since \( P(\text{win } y) \) is the probability of picking (1, 1), which is \(\frac{1}{6}\): \[ 0.3333 \times 2 + 0.6667 \times -1 + \frac{1}{6} \times y = 0 \] Solving for \( y \): \[ 0.6666 - 0.6667 + 0.1667y = 0 \] \[ 0.1667y = 0 - (0.6667 - 0.6666) \] \[ 0.1667y = 0.0001 \] \[ y = \frac{0.0001}{0.1667} \approx 0.0 \] Therefore, \( y = 0 \) is the value that would make the game fair. Miguel would neither be expected to win nor lose on average. In a complete sentence, "To make the game fair, the amount Miguel would win if he picks two chips with the number 1 should be $[/tex]0."
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