Correct Answer - A
In {1,2,3,……,11} there are 5 even numbers and 6 odd numbers . The sum even is possible only when both are odd or both are even .
Let A be the event that denotes both numbers are even and B be the event that denotes sum of numbers is even . Then , `n(A)=""^(5)C_(2)+""^(6)C_(2)` .
Required probability
`P(A//B)=(P(A cap B))/(P(B)) =(""^(5)C_(2)//""^(11)C_(2))/(((""^(6)C_(2)+""^(5)C_(2)))/(""^(11)C_(2)))=(""^(5)C_(2))/(""^(6)C_(2)+""^(5)C_(2))=(10)/(15+10)=(2)/(5)`