To solve the problem, we need to find the probability distribution of the random variable \( X \), which represents the maximum of the two numbers shown on a pair of rolled dice. We will also calculate the mean of \( X \).
### Step 1: Identify Possible Values of \( X \)
When rolling two dice, the possible values for \( X \) (the maximum number shown) can be 1, 2, 3, 4, 5, or 6.
### Step 2: Calculate the Probability for Each Value of \( X \)
1. **For \( X = 1 \)**:
- The only combination is (1, 1).
- Probability: \( P(X = 1) = \frac{1}{36} \).
2. **For \( X = 2 \)**:
- The combinations are (1, 2), (2, 1), and (2, 2).
- Probability: \( P(X = 2) = \frac{3}{36} = \frac{1}{12} \).
3. **For \( X = 3 \)**:
- The combinations are (1, 3), (2, 3), (3, 1), (3, 2), and (3, 3).
- Probability: \( P(X = 3) = \frac{5}{36} \).
4. **For \( X = 4 \)**:
- The combinations are (1, 4), (2, 4), (3, 4), (4, 1), (4, 2), (4, 3), and (4, 4).
- Probability: \( P(X = 4) = \frac{7}{36} \).
5. **For \( X = 5 \)**:
- The combinations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), and (5, 5).
- Probability: \( P(X = 5) = \frac{9}{36} = \frac{1}{4} \).
6. **For \( X = 6 \)**:
- The combinations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), and (6, 6).
- Probability: \( P(X = 6) = \frac{11}{36} \).
### Step 3: Construct the Probability Distribution Table
| \( X \) | Probability \( P(X) \) |
|---------|------------------------|
| 1 | \( \frac{1}{36} \) |
| 2 | \( \frac{3}{36} \) |
| 3 | \( \frac{5}{36} \) |
| 4 | \( \frac{7}{36} \) |
| 5 | \( \frac{9}{36} \) |
| 6 | \( \frac{11}{36} \) |
### Step 4: Calculate the Mean of \( X \)
The mean (expected value) \( E(X) \) is calculated using the formula:
\[
E(X) = \sum (X_i \cdot P(X_i))
\]
Calculating each term:
\[
E(X) = 1 \cdot \frac{1}{36} + 2 \cdot \frac{3}{36} + 3 \cdot \frac{5}{36} + 4 \cdot \frac{7}{36} + 5 \cdot \frac{9}{36} + 6 \cdot \frac{11}{36}
\]
Calculating each product:
- \( 1 \cdot \frac{1}{36} = \frac{1}{36} \)
- \( 2 \cdot \frac{3}{36} = \frac{6}{36} \)
- \( 3 \cdot \frac{5}{36} = \frac{15}{36} \)
- \( 4 \cdot \frac{7}{36} = \frac{28}{36} \)
- \( 5 \cdot \frac{9}{36} = \frac{45}{36} \)
- \( 6 \cdot \frac{11}{36} = \frac{66}{36} \)
Now, summing these values:
\[
E(X) = \frac{1 + 6 + 15 + 28 + 45 + 66}{36} = \frac{161}{36} \approx 4.47
\]
### Final Results
- **Probability Distribution Table**:
- \( P(X = 1) = \frac{1}{36} \)
- \( P(X = 2) = \frac{3}{36} \)
- \( P(X = 3) = \frac{5}{36} \)
- \( P(X = 4) = \frac{7}{36} \)
- \( P(X = 5) = \frac{9}{36} \)
- \( P(X = 6) = \frac{11}{36} \)
- **Mean of \( X \)**: \( E(X) \approx 4.47 \)
