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].
