Answer :
In this problem, Miguel is playing a game with a box containing four chips numbered [tex]\([1, 1, 3, 5]\)[/tex]. He selects two chips, and depending on the numbers, he either wins [tex]$\$[/tex]2[tex]$ (if both numbers are the same) or loses $[/tex]\[tex]$1$[/tex] (if the numbers are different).
Let's solve the problem step-by-step:
### Step 1: Determine the Possible Outcomes
First, we need to list all possible pairs of chips Miguel can choose. Since the total number of chips is 4 and he is choosing 2, we need to consider all combinations:
[tex]\[ \text{Possible pairs:} \][/tex]
[tex]\[ (1, 1), (1, 3), (1, 5), (1, 3), (1, 5), (3, 5) \][/tex]
### Step 2: Count Winning and Losing Outcomes
- Winning outcomes (both chips are the same): There is only one combination where both numbers are the same: [tex]\((1, 1)\)[/tex].
- Losing outcomes (both chips are different): The other five combinations have different numbers: [tex]\((1, 3), (1, 5), (1, 3), (1, 5), (3, 5)\)[/tex].
### Step 3: Fill Out the Table
The table we need to complete includes:
- [tex]\( x_t \)[/tex] which represents the different net amounts Miguel can win or lose.
- Counts of each type of outcome.
- The probabilities associated with each outcome.
From our analysis:
- Miguel wins [tex]$\$[/tex]2[tex]$ if he picks the pair \((1, 1)\). - Miguel loses $[/tex]\[tex]$1$[/tex] for all other combinations.
### Step 4: Total Outcomes and Counts
- Total Possible Outcomes: There are 6 possible pairs.
- Count of Winning Outcomes: 1 (for the pair [tex]\((1, 1)\)[/tex]).
- Count of Losing Outcomes: 5 (for the pairs [tex]\((1, 3), (1, 5), (1, 3), (1, 5), (3, 5)\)[/tex]).
### Step 5: Calculate Probabilities
- Probability of winning: [tex]\( P(\text{win}) = \frac{\text{number of winning outcomes}}{\text{total outcomes}} = \frac{1}{6} = 0.16666666666666666 \)[/tex]
- Probability of losing: [tex]\( P(\text{lose}) = \frac{\text{number of losing outcomes}}{\text{total outcomes}} = \frac{5}{6} = 0.8333333333333334 \)[/tex]
### Final Step: Fill All Values in the Table
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline $x_t$ & & 2 & & -1 \\ \hline & 1 & & 5 & $\because$ \\ \hline $P(x)$ & & 0.16666666666666666 & & 0.8333333333333334 \\ \hline \end{tabular} \][/tex]
So, the detailed solution is as follows:
- The amount Miguel will win if he draws two matching numbers is [tex]$\$[/tex]2[tex]$. - The amount Miguel will lose if he draws two different numbers is $[/tex]\[tex]$1$[/tex].
- There is 1 way to win and 5 ways to lose.
- The probability of winning is [tex]\( \frac{1}{6} = 0.16666666666666666 \)[/tex].
- The probability of losing is [tex]\( \frac{5}{6} = 0.8333333333333334 \)[/tex].
Let's solve the problem step-by-step:
### Step 1: Determine the Possible Outcomes
First, we need to list all possible pairs of chips Miguel can choose. Since the total number of chips is 4 and he is choosing 2, we need to consider all combinations:
[tex]\[ \text{Possible pairs:} \][/tex]
[tex]\[ (1, 1), (1, 3), (1, 5), (1, 3), (1, 5), (3, 5) \][/tex]
### Step 2: Count Winning and Losing Outcomes
- Winning outcomes (both chips are the same): There is only one combination where both numbers are the same: [tex]\((1, 1)\)[/tex].
- Losing outcomes (both chips are different): The other five combinations have different numbers: [tex]\((1, 3), (1, 5), (1, 3), (1, 5), (3, 5)\)[/tex].
### Step 3: Fill Out the Table
The table we need to complete includes:
- [tex]\( x_t \)[/tex] which represents the different net amounts Miguel can win or lose.
- Counts of each type of outcome.
- The probabilities associated with each outcome.
From our analysis:
- Miguel wins [tex]$\$[/tex]2[tex]$ if he picks the pair \((1, 1)\). - Miguel loses $[/tex]\[tex]$1$[/tex] for all other combinations.
### Step 4: Total Outcomes and Counts
- Total Possible Outcomes: There are 6 possible pairs.
- Count of Winning Outcomes: 1 (for the pair [tex]\((1, 1)\)[/tex]).
- Count of Losing Outcomes: 5 (for the pairs [tex]\((1, 3), (1, 5), (1, 3), (1, 5), (3, 5)\)[/tex]).
### Step 5: Calculate Probabilities
- Probability of winning: [tex]\( P(\text{win}) = \frac{\text{number of winning outcomes}}{\text{total outcomes}} = \frac{1}{6} = 0.16666666666666666 \)[/tex]
- Probability of losing: [tex]\( P(\text{lose}) = \frac{\text{number of losing outcomes}}{\text{total outcomes}} = \frac{5}{6} = 0.8333333333333334 \)[/tex]
### Final Step: Fill All Values in the Table
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline $x_t$ & & 2 & & -1 \\ \hline & 1 & & 5 & $\because$ \\ \hline $P(x)$ & & 0.16666666666666666 & & 0.8333333333333334 \\ \hline \end{tabular} \][/tex]
So, the detailed solution is as follows:
- The amount Miguel will win if he draws two matching numbers is [tex]$\$[/tex]2[tex]$. - The amount Miguel will lose if he draws two different numbers is $[/tex]\[tex]$1$[/tex].
- There is 1 way to win and 5 ways to lose.
- The probability of winning is [tex]\( \frac{1}{6} = 0.16666666666666666 \)[/tex].
- The probability of losing is [tex]\( \frac{5}{6} = 0.8333333333333334 \)[/tex].
Answer:
Step-by-step explanation:
To complete the table for Miguel's game, we need to determine the probability of each outcome and the total number of outcomes.
Here is a step-by-step approach:
1. **Determine All Possible Outcomes:**
- **Total Chips and Combinations:**
Miguel has 4 chips: two with number 1, one with number 3, and one with number 5.
We need to choose 2 chips out of these 4, and the number of possible combinations is calculated using the combination formula \( \binom{4}{2} \):
\[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
\]
- **List of Combinations and Outcomes:**
1. (1, 1) → Win $2
2. (1, 3) → Lose $1
3. (1, 5) → Lose $1
4. (3, 5) → Lose $1
5. (1, 3) → Lose $1
6. (1, 5) → Lose $1
**Note:** (1, 3) and (1, 5) are repeated, so there are a total of 6 unique combinations.
2. **Count the Frequency of Each Outcome:**
- **Winning $2:**
- (1, 1) occurs once.
- **Losing $1:**
- (1, 3) occurs twice.
- (1, 5) occurs twice.
- (3, 5) occurs once.
- Total losses: 2 (from 1, 3) + 2 (from 1, 5) + 1 (from 3, 5) = 5 times
3. **Calculate the Probability of Each Outcome:**
- **Probability of Winning $2:**
\[
P(2) = \frac{1}{6}
\]
- **Probability of Losing $1:**
\[
P(-1) = \frac{5}{6}
\]
4. **Fill in the Table:**
Based on the counts and probabilities:
\[
\begin{array}{|c|c|c|c|c|}
\hline
x_t & -1 & 2 \\
\hline
33 & 5 & 1 \\
\hline
P(x) & \frac{5}{6} & \frac{1}{6} \\
\hline
\end{array}
\]
**Explanation:**
- **\( x_t \)**: Represents the amount of money Miguel wins or loses.
- **Frequency**: The number of times each amount occurs.
- **Probability**: The likelihood of each outcome.
Thus, the completed table shows the frequencies and probabilities for each possible amount of money Miguel can win or lose.
