Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, one chip has the number 3, and the other chip has the number 5. Miguel must choose two chips, and if both chips have the same number, he wins \[tex]$2. If the two chips he chooses have different numbers, 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.)
[tex]\[
\begin{tabular}{|c|c|c|}
\hline
\( X_i \) & 2 & -1 \\
\hline
\( P(X_i) \) & & \\
\hline
\end{tabular}
\][/tex]

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



Answer :

Let's break down the solution to the question step-by-step.

### Part (a): Fill out the missing values in the table

We need to determine the probabilities associated with Miguel winning and losing the game. Let's start by understanding the scenarios.

Possible outcomes:
1. Miguel wins [tex]$2 if he draws two chips with the same number. 2. Miguel loses $[/tex]1 if he draws two chips with different numbers.

Chips in the box:
- Two chips with number 1.
- One chip with number 3.
- One chip with number 5.

Total combinations of drawing two chips:
Since Miguel is drawing 2 out of 4 chips, there are a total number of combinations given by the combination formula [tex]\( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)[/tex]. Here,
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \][/tex]

So, there are 6 possible pairs of chips Miguel can draw.

Winning combinations:
- (1, 1)

Since there is only one pair of chips that have the same number, Miguel wins [tex]$2 in this combination. Losing combinations: The losing combinations are the pairs of chips with different numbers, which include: - (1, 3) - (1, 5) - (1, 3) - (1, 5) - (3, 5) There are 5 different pairs where the numbers are different, meaning Miguel loses $[/tex]1 in these combinations.

Probabilities:

1. Probability of winning ([tex]$2): \[ P(X=2) = \frac{\text{Number of winning combinations}}{\text{Total number of combinations}} = \frac{1}{6} \approx 0.1667 \] 2. Probability of losing (-$[/tex]1):
[tex]\[ P(X=-1) = \frac{\text{Number of losing combinations}}{\text{Total number of combinations}} = \frac{5}{6} \approx 0.8333 \][/tex]

Thus, the completed probability table is:

\begin{tabular}{|c|c|c|}
\hline[tex]$X_i$[/tex] & 2 & -1 \\
\hline[tex]$P\left(x_i\right)$[/tex] & 0.1667 & 0.8333 \\
\hline
\end{tabular}

### Part (b): Expected value from playing the game

The expected value [tex]\(E(X)\)[/tex] represents the average amount of money Miguel can expect to win or lose per game in the long run. It is calculated using the formula:

[tex]\[ E(X) = \sum (X_i \times P(X_i)) \][/tex]

Using the values from the table:

1. For [tex]\(X = 2\)[/tex]:
[tex]\[ 2 \times 0.1667 = 0.3334 \][/tex]

2. For [tex]\(X = -1\)[/tex]:
[tex]\[ -1 \times 0.8333 = -0.8333 \][/tex]

Now, summing these products, we get the expected value:

[tex]\[ E(X) = 0.3334 + (-0.8333) = -0.5 \][/tex]

Thus, the expected value from playing the game is [tex]\(-\$0.50\)[/tex].

To summarize:

(a) The completed probability table is:
\begin{tabular}{|c|c|c|}
\hline[tex]$X_i$[/tex] & 2 & -1 \\
\hline[tex]$P\left(x_i\right)$[/tex] & 0.1667 & 0.8333 \\
\hline
\end{tabular}

(b) Miguel's expected value from playing the game is [tex]\(-\$0.50\)[/tex].