Two numbers are selected simultaneously from the set {6, 7, 8, 9, ……, 39}. If the sum of selected numbers is even then the probability that both the selected numbers are odd, is equal to

$\frac{1}{2}$

The correct answer is Option 4: $\frac{1}{2}$

Total numbers from 6 to 39 = 34 numbers

• Even numbers = 17
• Odd numbers = 17

A : Sum of selected numbers is even.

B : Selected numbers are odd.

$P(A)=\frac{{ }^{17} C_2+{ }^{17} C_2}{{ }^{34} C_2}, P(A \cap B)=\frac{{ }^{17} C_2}{{ }^{34} C_2}$

$P(B ~|~ A)=\frac{P(A \cap B)}{P(A)}=\frac{{ }^{17} C_2}{{ }^{17} C_2+{ }^{17} C_2}$

=$\frac{136}{272}$

=$\frac{1}{2}$