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Two integers are selected at random from the set $\{1,2, \ldots, 11\}$. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :
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The correct answer is:
$\frac{2}{5}$
Probability of getting sum of selected two numbers is even
$=P\left(E_{1}\right)=\frac{{ }^{5} C_{2}+{ }^{5} C_{2}}{{ }^{11} C_{2}}$
Probability of getting sum is even and selected numbers are also even $P\left(E_{2}\right)=\frac{{ }^{5} C_{2}}{{ }^{11} C_{2}}$
Hence, $P\left(\frac{E_{2}}{E_{1}}\right)=\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}+{ }^{5} C_{2}}=\frac{10}{15+10}=\frac{2}{5}$.
$=P\left(E_{1}\right)=\frac{{ }^{5} C_{2}+{ }^{5} C_{2}}{{ }^{11} C_{2}}$
Probability of getting sum is even and selected numbers are also even $P\left(E_{2}\right)=\frac{{ }^{5} C_{2}}{{ }^{11} C_{2}}$
Hence, $P\left(\frac{E_{2}}{E_{1}}\right)=\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}+{ }^{5} C_{2}}=\frac{10}{15+10}=\frac{2}{5}$.
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