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

  Sum even = both even or both odd

$n\left(A\right)=5_{C_2}+6_{C_2}$

$n(A\cap B)=6_{C_2}$

$P\left(B/A\right)=\frac{n(A\cap B)}{n\left(A\right)}$

                   $=\frac{6_{C_2}}{5_{C_2}+6_{C_2}}$

                   $=\frac{15}{10+15}=\frac{3}{5}$

Required Probability is  $\frac{3}{5}$