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

We have the finite set $$\{1,2,3,\ldots ,11\}$$ from which two distinct integers are chosen at random, each unordered pair being equally likely.

The total of the two selected numbers is already known to be even. An even sum can arise in exactly two ways: either both chosen numbers are even or both are odd. Our task is to find the probability that the first of these possibilities occurs, under the condition that the sum is even.

Let us list the parity composition of the set. The even numbers are $$2,4,6,8,10,$$ so there are $$5$$ evens. The odd numbers are $$1,3,5,7,9,11,$$ so there are $$6$$ odds.

Number of unordered pairs where both numbers are even:

First, we state the combination formula: choosing $$r$$ objects from $$n$$ distinct objects is counted by $$\displaystyle \binom{n}{r}=\frac{n!}{r!(n-r)!}.$$ Using this, the count of pairs of evens is

$$\binom{5}{2}=\frac{5!}{2!\,3!}=10.$$

Number of unordered pairs where both numbers are odd:

$$\binom{6}{2}=\frac{6!}{2!\,4!}=15.$$

Because an even sum must be obtained from either two evens or two odds, every pair with even sum has now been enumerated. Hence the total number of unordered pairs whose sum is even equals

$$10+15=25.$$

Now we invoke the definition of conditional probability. If event $$E$$ denotes “sum is even” and event $$A$$ denotes “both numbers are even,” then

$$P(A\mid E)=\frac{\text{Number of pairs in }A}{\text{Number of pairs in }E} =\frac{10}{25} =\frac{2}{5}.$$

So the conditional probability that both selected numbers are even, given that their sum is even, equals $$\dfrac{2}{5}.$$

Hence, the correct answer is Option C.