** Answer: (a) \(\frac{5}{8}\)**

In the set of integers from 1 to 9, there are **four even** numbers 2, 4, 6, 8, and **5 odd** numbers 1, 3, 5, 7, 9.

Let A: Event of choosing odd numbers ⇒ n(A) = ^{5}C_{2}

(∵ 2 numbers are chosen from 5 odd numbers)

B: Event of getting the sum as an even number. ⇒ n(B) = ^{4}C_{2} + ^{5}C_{2}

(∵ The sum is even if both the numbers are chosen are even or both are odd)

∴ n(A ∩ B) = ^{5}C_{2}

**(Event of getting sum as even if both the numbers are odd) **

∴ P(Selecting both odd numbers or getting an even sum)

**= P(A/B) = \(\frac{P(A\,\cap\,B)}{P(A)}\) = \(\frac{n(A\,\cap\,B)}{n(B)}\) = \(\frac{^5C_2}{^4C_2+^5C_2}\) = \(\frac{10}{16} =\frac{5}{8}\)**