Two integers are selected at random from the integers 1 to 9. If their sum is even, then find the probability that both integers are odd.

Explanation

To solve the problem step by step, we will find the probability that both selected integers are odd given that their sum is even.

### Step 1: Identify the Sample Space
The integers we can choose from are 1 to 9. Therefore, the sample space \( S \) consists of the integers:
\[ S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \]

### Step 2: Determine the Total Number of Possible Pairs
We need to select 2 integers from the 9 available. The total number of ways to choose 2 integers from 9 is given by the combination formula:
\[ \text{Total pairs} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36 \]

### Step 3: Define Events A and B
- Let \( A \) be the event that the sum of the two selected integers is even.
- Let \( B \) be the event that both integers are odd.

### Step 4: Find the Event A (Sum is Even)
The sum of two integers is even if:
1. Both integers are even.
2. Both integers are odd.

**Odd integers from 1 to 9:** \( \{1, 3, 5, 7, 9\} \) (5 odd integers)
**Even integers from 1 to 9:** \( \{2, 4, 6, 8\} \) (4 even integers)

**Pairs of odd integers (sum is even):**
- The number of ways to choose 2 odd integers from 5:
\[ \text{Odd pairs} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \]

**Pairs of even integers (sum is even):**
- The number of ways to choose 2 even integers from 4:
\[ \text{Even pairs} = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \]

**Total pairs in event A:**
\[ |A| = \text{Odd pairs} + \text{Even pairs} = 10 + 6 = 16 \]

### Step 5: Find the Event B (Both Integers are Odd)
The pairs where both integers are odd (event B) are:
- \( (1, 3), (1, 5), (1, 7), (1, 9) \)
- \( (3, 5), (3, 7), (3, 9) \)
- \( (5, 7), (5, 9) \)
- \( (7, 9) \)

**Total pairs in event B:**
\[ |B| = 10 \text{ (as calculated in step 4)} \]

### Step 6: Find the Intersection of Events A and B
Since all pairs in event B (both integers are odd) also result in an even sum, we have:
\[ A \cap B = B \]
Thus, \( |A \cap B| = |B| = 10 \).

### Step 7: Calculate the Probability \( P(B|A) \)
Using the conditional probability formula:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{|A \cap B|}{|A|} \]
Substituting the values we found:
\[ P(B|A) = \frac{|B|}{|A|} = \frac{10}{16} = \frac{5}{8} \]

### Final Answer
The probability that both integers are odd given that their sum is even is:
\[ \frac{5}{8} \]