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

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The integers from 1 through 11 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. Out of these, there are 5 even and 6 odd integers. 

Let A: Both numbers chosen are odd 

B: Sum of numbers is even at random 

S: Choosing 2 numbers from 11 numbers. 

Then, n(S) = 11C2 

n(A) = 6C2 (∴  There are 6 odd integers) 

As the sum of both chosen integers can be even if both are even or both are odd, so 

n(B) = 6C2 + 5C2 

and n(A ∩ B) = 6C2 

∴                               P(A) = \(\frac{n(A)}{n(S)} = \frac {^6C_2}{^{11}C_2}= \frac{6\times 5}{11 \times 10} =\frac{3}{11}\)

  P(B) = \(\frac{n(B)}{n(S)} = \frac {^6C_2+^5C_2}{^{11}C_2}= \frac{6\times 5+ 5\times 4}{11 \times 10} =\frac{5}{11}\)

\(P(A\,\cap\,B) = \frac{n(A\,\cap\,B)}{n(S)} =\frac{^6C_2}{^{11}C_2} =\frac{5}{11}\)

∴ P(A/B) = \(\frac {P(A\,\cap\,B)}{P(B)}=\frac{\frac{3}{11}}{\frac{5}{11}} =\frac{3}{5}\)

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