# Two numbers are selected at random from the integer 1 through 9. If the sum is even, find the probability that both numbers are odd.

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Two numbers are selected at random from the integer 1 through 9. If the sum is even, find the probability that both numbers are odd.

(a) $\frac{5}{8}$

(b) $\frac{3}{8}$

(c) $\frac{3}{10}$

(d) None of these

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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) = 5C2

(∵ 2 numbers are chosen from 5 odd numbers)

B: Event of getting the sum as an even number. ⇒ n(B) = 4C2 + 5C2

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

∴ n(A ∩ B) = 5C2

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