Two integers are selected at random from the integers 1 to 11. If their sum is even find the probability that both integers are odd.

Explanation

To solve the problem step by step, we will follow the logical reasoning presented in the video transcript.

### Step 1: Determine the Total Sample Space
We need to select 2 integers from the integers 1 to 11. The total number of ways to select 2 integers from 11 can be calculated using the combination formula:

\[ \text{Total ways} = \binom{11}{2} = \frac{11 \times 10}{2} = 55 \]

**Hint for Step 1:** Remember that the combination formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) is used to find the number of ways to choose \(r\) items from \(n\) items without regard to the order of selection.

### Step 2: Define the Event of Interest
Let \(E\) be the event that the sum of the two selected integers is even. The sum of two integers is even if both integers are even or both integers are odd.

### Step 3: Count the Even and Odd Integers
From the integers 1 to 11:
- The even integers are: 2, 4, 6, 8, 10 (Total = 5 even integers)
- The odd integers are: 1, 3, 5, 7, 9, 11 (Total = 6 odd integers)

### Step 4: Calculate the Number of Ways for Each Case
1. **Both integers are even:**
The number of ways to choose 2 even integers from the 5 available is:

\[ \text{Ways (even)} = \binom{5}{2} = \frac{5 \times 4}{2} = 10 \]

2. **Both integers are odd:**
The number of ways to choose 2 odd integers from the 6 available is:

\[ \text{Ways (odd)} = \binom{6}{2} = \frac{6 \times 5}{2} = 15 \]

### Step 5: Total Number of Favorable Outcomes for Event \(E\)
The total number of ways to select 2 integers such that their sum is even (event \(E\)) is the sum of both cases:

\[ \text{Total ways for } E = \text{Ways (even)} + \text{Ways (odd)} = 10 + 15 = 25 \]

### Step 6: Define Event \(F\)
Let \(F\) be the event that both integers are odd. We have already calculated that the number of ways to choose 2 odd integers is:

\[ \text{Number of elements in } F = 15 \]

### Step 7: Calculate the Probability of Event \(F\) Given Event \(E\)
We want to find the conditional probability \(P(F|E)\):

\[ P(F|E) = \frac{P(F \cap E)}{P(E)} \]

Where:
- \(P(F \cap E)\) is the number of ways both integers are odd (which is 15).
- \(P(E)\) is the total number of ways to get an even sum (which is 25).

Thus, the probability is:

\[ P(F|E) = \frac{15}{25} = \frac{3}{5} \]

### Final Answer
The probability that both integers are odd given that their sum is even is:

\[ \frac{3}{5} \]

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