1. Identify odd and even numbers in $\{1,2,\dots,11\}$

From $1$ to $11$:

• Even numbers: $\{2,4,6,8,10\}$, so there are $5$ even numbers.
• Odd numbers: $\{1,3,5,7,9,11\}$, so there are $6$ odd numbers.

2. Condition for the sum to be even

The sum of two selected integers is even if:

• both are even, or
• both are odd.

So we need: [ P(\text{both even} ,|, \text{sum even}) = \frac{P(\text{both even})}{P(\text{sum even})} ] Since selection is uniform, we can count favorable outcomes directly.

3. Count ways to select two integers

Since two integers are selected from the set, we interpret this as choosing 2 distinct numbers.

• Number of ways to choose 2 even numbers: [ \binom{5}{2} = 10 ]

• Number of ways to choose 2 odd numbers: [ \binom{6}{2} = 15 ]

Thus, number of selections with even sum: [ 10 + 15 = 25 ]

4. Compute the conditional probability

Favorable cases for “both numbers are even” under the condition “sum is even”: [ \binom{5}{2} = 10 ]

Therefore, [ P(\text{both even} ,|, \text{sum even}) = \frac{10}{25} = \frac{2}{5} ]

5. Match with the options

[ \frac{2}{5} ] So the correct option is A.