Two integers are selected at random from integers 1 to 11. If the sum is even, the probability that both the numbers are odd, is

$\frac{3}{5}$

Out of integers from 1 to 11, there are 5 even integers and 6 odd integers.

Consider the following events:

A = Both the numbers chosen are odd,

B = The sum of the numbers chosen is even.

∴ Required probability

$ = P( A/ B)$

= Probability that the two numbers chosen are odd if it is given that the sum of the numbers chosen is even.

$=\frac{^C_2}{^C_2+ {^6C}_2}=\frac{3}{5}$