Two integers are selected at random from the set 1, 2, …., 11. Given that the sum of the selected numbers is even, the conditional probability that both the numbers are even is:
(a) $\dfrac{2}{5}$,
(b) $\dfrac{1}{2}$,
(c) $\dfrac{3}{5}$,
(d) $\dfrac{7}{10}$.

Hint: We start solving the problem by recalling the fact that the sum of two numbers is even if both the numbers are odd or both the numbers are even. We then find the total no. of ways of choosing two numbers from given 11 numbers, choosing two odd numbers from given 11 numbers and choosing two even numbers from given 11 numbers. We then find the probabilities of choosing two numbers from given 11 numbers, choosing two odd numbers from given 11 numbers and choosing two even numbers from given 11 numbers. After finding these values we find the conditional probability required.

Complete step by step answer:
According to the problem, we are selecting two integers randomly from the set 1, 2, 3, ……,11. We need to find the conditional probability that the both selected numbers are even, given that the sum of two selected numbers is even.
Let us first find the probability that the sum of the two selected numbers are even. We know that the sum of two numbers is even if both the numbers are odd or both the numbers are even.
Let us $P\left( S \right)$ be the probability that the sum of two selected numbers is even, $P\left( {{S}_{E}} \right)$ be the probability that the both selected numbers are even and $P\left( {{S}_{O}} \right)$ be the probability that both selected numbers are odd.
We know that the no. of ways to select two numbers from the given 11 numbers is ${}^{11}{{C}_{2}}$. We also know that the no. of ways to select two odd numbers from the given 11 numbers (they has 6 odd numbers) is ${}^{6}{{C}_{2}}$ and the no. of ways to select two even numbers from the given 11 numbers (they has 5 even numbers) is ${}^{5}{{C}_{2}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Let us find the probability that the two selected numbers are odd.
So, we have $P\left( {{S}_{O}} \right)=\dfrac{\text{No}\text{. of ways to select two odd numbers from 11 numbers}}{\text{No}\text{. of ways to select two numbers from 11 numbers}}$.
$\Rightarrow P\left( {{S}_{O}} \right)=\dfrac{{}^{6}{{C}_{2}}}{{}^{11}{{C}_{2}}}$.
$\Rightarrow P\left( {{S}_{0}} \right)=\dfrac{\dfrac{6!}{2!\left( 6-2 \right)!}}{\dfrac{11!}{2!\left( 11-2 \right)!}}$.
$\Rightarrow P\left( {{S}_{0}} \right)=\dfrac{\dfrac{6!}{4!}}{\dfrac{11!}{9!}}$.
$\Rightarrow P\left( {{S}_{0}} \right)=\dfrac{5\times 6}{11\times 10}$.
$\Rightarrow P\left( {{S}_{0}} \right)=\dfrac{3}{11}$.
Let us find the probability that the two selected numbers are even.
So, we have $P\left( {{S}_{E}} \right)=\dfrac{\text{No}\text{. of ways to select two even numbers from 11 numbers}}{\text{No}\text{. of ways to select two numbers from 11 numbers}}$.
$\Rightarrow P\left( {{S}_{E}} \right)=\dfrac{{}^{5}{{C}_{2}}}{{}^{11}{{C}_{2}}}$.
$\Rightarrow P\left( {{S}_{E}} \right)=\dfrac{\dfrac{5!}{2!\left( 5-2 \right)!}}{\dfrac{11!}{2!\left( 11-2 \right)!}}$.
$\Rightarrow P\left( {{S}_{E}} \right)=\dfrac{\dfrac{5!}{3!}}{\dfrac{11!}{9!}}$.
$\Rightarrow P\left( {{S}_{0}} \right)=\dfrac{5\times 4}{11\times 10}$.
$\Rightarrow P\left( {{S}_{0}} \right)=\dfrac{2}{11}$.
Let us find the probability that the sum of the two selected numbers is even.
So, we have $P\left( S \right)=P\left( {{S}_{0}} \right)+P\left( {{S}_{E}} \right)$.
$\Rightarrow P\left( S \right)=\dfrac{3}{11}+\dfrac{2}{11}$.
$\Rightarrow P\left( S \right)=\dfrac{5}{11}$.
Now, we find the conditional probability that the both selected numbers are even, given that the sum of two selected numbers is even.
So, we have to find $\dfrac{P\left( {{S}_{E}} \right)}{P\left( S \right)}$.
$\Rightarrow \dfrac{P\left( {{S}_{E}} \right)}{P\left( S \right)}=\dfrac{\dfrac{2}{11}}{\dfrac{5}{11}}$.
$\Rightarrow \dfrac{P\left( {{S}_{E}} \right)}{P\left( S \right)}=\dfrac{2}{5}$.
We have found the conditional probability that the both selected numbers are even, given that the sum of two selected numbers is even as $\dfrac{2}{5}$.
∴ The conditional probability that the both selected numbers are even, given that the sum of two selected numbers is even is $\dfrac{2}{5}$.

So, the correct answer is “Option A”.

Note: In this problem while calculating the conditional probability, we have directly taken the value of $P\left( {{S}_{E}} \right)$ as the intersection of two events – sum of two selected numbers is even and selecting two even numbers gives selecting two even numbers. We check the answer by finding the condition probability with odd numbers and finding the sum of both probabilities as 1. Similarly, we can also expect problems to find conditional probability with respect to the sum of two selected numbers is odd.