Two integers are selected at random from the set $\left\{1,2,\mathrm{3.....11}\right\}$ . Given that the sum of selected numbers is even, then the conditional probability that both numbers are even is

$$\begin{gathered} \mathrm{Two} \mathrm{integers} \mathrm{are} \mathrm{selected} \mathrm{at} \mathrm{random} \mathrm{from} \mathrm{the} \mathrm{set} 1 , 2 , 3.....11; \\ \mathrm{A}=\mathrm{Sum} \mathrm{of} \mathrm{the} 2 \mathrm{integers} \mathrm{selected} \mathrm{is} \mathrm{even} \\ (\mathrm{both} \mathrm{even} \mathrm{or} \mathrm{both} \mathrm{odd}); \\ \Rightarrow \mathrm{n}\left(\mathrm{A}\right){=}^5{\mathrm{C}}_2{+}^6{\mathrm{C}}_2; \\ \mathrm{B}= \mathrm{selecting} \mathrm{two} \mathrm{even} \mathrm{integers}; \\ \Rightarrow \mathrm{n}(A\cap \mathrm{B}){=}^5{\mathrm{C}}_2; \\ now \mathrm{P}\left(\mathrm{B}/\mathrm{A}\right)=\frac{\mathrm{n}(\mathrm{A}\cap \mathrm{B})}{\mathrm{n}\left(\mathrm{A}\right)}=\frac{{ }^5{\mathrm{C}}_2}{{ }^5{\mathrm{C}}_2{+}^6{\mathrm{C}}_2}=\frac{10}{10+15}=\frac{10}{25}=\frac{2}{5} \end{gathered}$$