Two integers are selected at random from the set {1,2,…………..,11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is

Options

A. `(2)/(5)`
B. `(1)/(2)`
C. `(7)/(10)`
D. `(3)/(5)`

Explanation

To solve the problem, we need to find the conditional probability that both selected integers are even given that their sum is even.

### Step-by-Step Solution:

1. **Identify the Set and Count the Even and Odd Integers**:
The set of integers is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.
- Even integers: {2, 4, 6, 8, 10} (5 even numbers)
- Odd integers: {1, 3, 5, 7, 9, 11} (6 odd numbers)

2. **Determine the Total Ways to Choose 2 Integers**:
The total number of ways to choose 2 integers from 11 is given by the combination formula:
\[ \text{Total ways} = \binom{11}{2} = \frac{11 \times 10}{2} = 55 \]

3. **Identify the Condition of the Sum Being Even**:
The sum of two integers is even if:
- Both integers are even, or
- Both integers are odd.

4. **Calculate the Number of Favorable Outcomes for Each Case**:
- **Case 1**: Both integers are even.
\[ \text{Ways to choose 2 even integers} = \binom{5}{2} = \frac{5 \times 4}{2} = 10 \]
- **Case 2**: Both integers are odd.
\[ \text{Ways to choose 2 odd integers} = \binom{6}{2} = \frac{6 \times 5}{2} = 15 \]

5. **Total Ways for the Sum to be Even**:
\[ \text{Total ways for sum to be even} = \text{Ways (even)} + \text{Ways (odd)} = 10 + 15 = 25 \]

6. **Calculate the Conditional Probability**:
We need to find the probability that both integers are even given that their sum is even. This is given by:
\[ P(\text{Both even} | \text{Sum is even}) = \frac{P(\text{Both even and sum is even})}{P(\text{Sum is even})} \]
Here, \(P(\text{Both even and sum is even})\) is simply the number of ways to choose both even integers (10), and \(P(\text{Sum is even})\) is the total ways for the sum to be even (25).

Thus, we have:
\[ P(\text{Both even} | \text{Sum is even}) = \frac{10}{25} = \frac{2}{5} \]

### Final Answer:
The conditional probability that both selected numbers are even given that their sum is even is \(\frac{2}{5}\).

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