Sure, let's work through this problem step by step.
First, identify the possible outcomes when Miguel picks two chips from the box. Given four chips numbered as follows:
- Two chips with the number 1
- One chip with the number 3
- One chip with the number 5
Miguel can pick any two chips from these. Let's list the possible pairs:
1. (1, 1)
2. (1, 1) - Since there are two chips labeled 1, this combination counts twice.
3. (1, 3)
4. (1, 5)
5. (3, 5)
6. (3, 1) - But since we are only choosing, this is same as combinations above.
There are a total of 6 possible outcomes.
Next, we need to categorize these outcomes into winning or losing scenarios and assign the respective probabilities:
- Miguel wins [tex]$\$[/tex]2[tex]$ if both selected chips have the same number.
- Winning pairs: (1, 1) and (1, 1) (counted twice), (3, 3) (but there is no chip 3 twice so doesn’t count)
- Miguel loses $[/tex]\[tex]$1$[/tex] if the selected chips have different numbers.
- Losing pairs: (1, 3), (1, 5), (3, 5)
There are 3 winning outcomes where Miguel wins [tex]$\$[/tex]2[tex]$ and 3 losing outcomes where Miguel loses $[/tex]\[tex]$1$[/tex]. This tells us:
- The probability of winning [tex]$\$[/tex]2[tex]$ (P(X = 2)) is \(\frac{3}{6} = 0.5\).
- The probability of losing $[/tex]\[tex]$1$[/tex] (P(X = -1)) is [tex]\(\frac{3}{6} = 0.5\)[/tex].
Now, fill in the missing values in the table:
\begin{tabular}{|c|c|c|}
\hline
[tex]$X_i$[/tex] & 2 & -1 \\
\hline
[tex]$P\left(x_i\right)$[/tex] & 0.5 & 0.5 \\
\hline
\end{tabular}
Therefore, the completed table should be as shown above:
- X_i = 2 when Miguel wins [tex]$2.
- P(X = 2) = 0.5
- X_i = -1 when Miguel loses $[/tex]1.
- P(X = -1) = 0.5
