To solve the problem, we will follow these steps:
### Step 1: Identify the Sample Space
We are selecting 2 numbers from the first six positive integers: {1, 2, 3, 4, 5, 6}. The total number of ways to select 2 numbers from 6 is given by the combination formula:
\[
\text{Total combinations} = \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
\]
### Step 2: Determine the Values of X
Let \( X \) denote the larger of the two selected numbers. The possible values of \( X \) can be 2, 3, 4, 5, or 6.
### Step 3: Calculate the Probability Distribution of X
We will find the probability for each value of \( X \):
- **For \( X = 2 \)**: The only pair is (1, 2).
- Combinations: 1
- Probability: \( P(X=2) = \frac{1}{15} \)
- **For \( X = 3 \)**: The pairs are (1, 3), (2, 3).
- Combinations: 2
- Probability: \( P(X=3) = \frac{2}{15} \)
- **For \( X = 4 \)**: The pairs are (1, 4), (2, 4), (3, 4).
- Combinations: 3
- Probability: \( P(X=4) = \frac{3}{15} = \frac{1}{5} \)
- **For \( X = 5 \)**: The pairs are (1, 5), (2, 5), (3, 5), (4, 5).
- Combinations: 4
- Probability: \( P(X=5) = \frac{4}{15} \)
- **For \( X = 6 \)**: The pairs are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6).
- Combinations: 5
- Probability: \( P(X=6) = \frac{5}{15} = \frac{1}{3} \)
### Step 4: Construct the Probability Distribution Table
\[
\begin{array}{|c|c|}
\hline
X & P(X) \\
\hline
2 & \frac{1}{15} \\
3 & \frac{2}{15} \\
4 & \frac{3}{15} \\
5 & \frac{4}{15} \\
6 & \frac{5}{15} \\
\hline
\end{array}
\]
### Step 5: Calculate the Mean of the Distribution
The mean \( \mu \) is calculated as follows:
\[
\mu = \sum (X \cdot P(X)) = 2 \cdot \frac{1}{15} + 3 \cdot \frac{2}{15} + 4 \cdot \frac{3}{15} + 5 \cdot \frac{4}{15} + 6 \cdot \frac{5}{15}
\]
Calculating each term:
\[
= \frac{2}{15} + \frac{6}{15} + \frac{12}{15} + \frac{20}{15} + \frac{30}{15}
\]
\[
= \frac{70}{15} = \frac{14}{3}
\]
### Step 6: Calculate the Variance of the Distribution
The variance \( \sigma^2 \) is calculated as follows:
\[
\sigma^2 = \sum (X - \mu)^2 \cdot P(X)
\]
Calculating each term:
\[
= (2 - \frac{14}{3})^2 \cdot \frac{1}{15} + (3 - \frac{14}{3})^2 \cdot \frac{2}{15} + (4 - \frac{14}{3})^2 \cdot \frac{3}{15} + (5 - \frac{14}{3})^2 \cdot \frac{4}{15} + (6 - \frac{14}{3})^2 \cdot \frac{5}{15}
\]
Calculating each squared difference and multiplying by the probabilities:
1. \( (2 - \frac{14}{3})^2 = \left(-\frac{8}{3}\right)^2 = \frac{64}{9} \)
2. \( (3 - \frac{14}{3})^2 = \left(-\frac{5}{3}\right)^2 = \frac{25}{9} \)
3. \( (4 - \frac{14}{3})^2 = \left(-\frac{2}{3}\right)^2 = \frac{4}{9} \)
4. \( (5 - \frac{14}{3})^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \)
5. \( (6 - \frac{14}{3})^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \)
Now substituting back into the variance formula:
\[
\sigma^2 = \frac{64}{9} \cdot \frac{1}{15} + \frac{25}{9} \cdot \frac{2}{15} + \frac{4}{9} \cdot \frac{3}{15} + \frac{1}{9} \cdot \frac{4}{15} + \frac{16}{9} \cdot \frac{5}{15}
\]
Calculating:
\[
= \frac{64}{135} + \frac{50}{135} + \frac{12}{135} + \frac{4}{135} + \frac{80}{135} = \frac{210}{135} = \frac{14}{9}
\]
### Final Answers
- **Probability Distribution of X**:
\[
\begin{array}{|c|c|}
\hline
X & P(X) \\
\hline
2 & \frac{1}{15} \\
3 & \frac{2}{15} \\
4 & \frac{3}{15} \\
5 & \frac{4}{15} \\
6 & \frac{5}{15} \\
\hline
\end{array}
\]
- **Mean**: \( \frac{14}{3} \)
- **Variance**: \( \frac{14}{9} \)
