Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).

Explanation

To find the expected value \( E(X) \) where \( X \) is the larger of two numbers selected at random without replacement from the first six positive integers (1, 2, 3, 4, 5, 6), we can follow these steps:

### Step 1: Determine the Sample Space
The total number of ways to select 2 numbers from 6 without replacement is given by:
\[ \text{Total ways} = 6 \times 5 = 30 \]

### Step 2: Identify Possible Values of \( X \)
The possible values for \( X \) (the larger of the two selected numbers) can be 2, 3, 4, 5, or 6.

### Step 3: Calculate the Probability for Each Value of \( X \)

1. **When \( X = 2 \)**:
- The only pair is (1, 2).
- Number of favorable outcomes = 1.
- Probability \( P(X = 2) = \frac{1}{30} \).

2. **When \( X = 3 \)**:
- Possible pairs: (1, 3), (2, 3).
- Number of favorable outcomes = 2.
- Probability \( P(X = 3) = \frac{2}{30} = \frac{1}{15} \).

3. **When \( X = 4 \)**:
- Possible pairs: (1, 4), (2, 4), (3, 4).
- Number of favorable outcomes = 3.
- Probability \( P(X = 4) = \frac{3}{30} = \frac{1}{10} \).

4. **When \( X = 5 \)**:
- Possible pairs: (1, 5), (2, 5), (3, 5), (4, 5).
- Number of favorable outcomes = 4.
- Probability \( P(X = 5) = \frac{4}{30} = \frac{2}{15} \).

5. **When \( X = 6 \)**:
- Possible pairs: (1, 6), (2, 6), (3, 6), (4, 6), (5, 6).
- Number of favorable outcomes = 5.
- Probability \( P(X = 6) = \frac{5}{30} = \frac{1}{6} \).

### Step 4: Create the Probability Distribution Table
\[ \begin{array}{|c|c|} \hline X & P(X) \\ \hline 2 & \frac{1}{30} \\ 3 & \frac{2}{30} = \frac{1}{15} \\ 4 & \frac{3}{30} = \frac{1}{10} \\ 5 & \frac{4}{30} = \frac{2}{15} \\ 6 & \frac{5}{30} = \frac{1}{6} \\ \hline \end{array} \]

### Step 5: Calculate the Expected Value \( E(X) \)
The expected value \( E(X) \) is calculated as follows:
\[ E(X) = \sum (X \cdot P(X)) \]
Calculating each term:
\[ E(X) = 2 \cdot \frac{1}{30} + 3 \cdot \frac{1}{15} + 4 \cdot \frac{1}{10} + 5 \cdot \frac{2}{15} + 6 \cdot \frac{1}{6} \]

Calculating each component:
- \( 2 \cdot \frac{1}{30} = \frac{2}{30} \)
- \( 3 \cdot \frac{1}{15} = \frac{3}{15} = \frac{6}{30} \)
- \( 4 \cdot \frac{1}{10} = \frac{4}{10} = \frac{12}{30} \)
- \( 5 \cdot \frac{2}{15} = \frac{10}{15} = \frac{20}{30} \)
- \( 6 \cdot \frac{1}{6} = 1 = \frac{30}{30} \)

Now summing these:
\[ E(X) = \frac{2 + 6 + 12 + 20 + 30}{30} = \frac{70}{30} = \frac{7}{3} \]

Thus, the expected value \( E(X) \) is:
\[ E(X) = \frac{14}{3} \]

### Final Answer
The expected value \( E(X) \) is \( \frac{14}{3} \).

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