The chips have the number 1. One chip contains four chips with numbers written on them. Miguel must choose two chips. One chip has the number 3, and the other chip has the number 5. If the two chips he chooses have different numbers, he wins \[tex]$2. If the two chips have the same number, he loses \$[/tex]1.

a. Let [tex]$X$[/tex] be the amount of money Miguel will receive or owe. Fill out the missing values in the table. (Hint: The total possible outcomes are six because there are four chips and you are choosing two of them.)

\begin{tabular}{|c|c|c|}
\hline
[tex]$x$[/tex] & 2 & -1 \\
\hline
[tex]$P(x)$[/tex] & & \\
\hline
\end{tabular}

b. What is Miguel's expected value from playing the game?

c. Based on the expected value in the previous step, how much money should Miguel expect to win or lose each time he plays?

d. What value should be assigned to choosing two chips with the number 1 to make the game fair? Explain your answer using a complete sentence and/or an equation.



Answer :

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."