What is the probability that both the numbers are odd with given conditions? 2 Numbers are selected at random from the integers 1 through 9.If the sum is even,find the probability that both the numbers are odd.


My approach:

A:Event of getting sum as even, B:Event of getting both the numbers as odd

p(B/A)=
I think to calculate the sum  as even we do like this 

(1,3),(1,5),(1,7),(1,9)

Similarly (2,2),(2,4),(2,6),(2,8) etc for 

3,1....4,2......5,1..6,2....,7,1..8,2.......9,1...

There are 28 such Numbers.

I think i am doing wrong because the Ans are not considering any such calculations and i am not aware of why i am doing wrong.

Expected Ans:5/8
There are five odd numbers and four even ones. Given that we drew either two odds or two evens, what is the probability that we drew two odds?

Assuming we drew with replacement:

Probability of two evens is $(4/9)^2$. Probability of two odds is $(5/9)^2$. Therefore the relative probability of two odds is $\frac{(5/9)^2}{(4/9)^2+(5/9)^2} = \frac{25}{16+25} = \frac{25}{41}$.

Assuming we drew without replacement:

Probability of two evens is $(4/9)(3/8)$. Probability of two odds is $(5/9)(4/8)$. Therefore the relative probability of two odds is $\frac{(5/9)(4/8)}{(4/9)(3/8) + (5/9)(4/8)} = \frac{20}{12+20} = \frac{10}{16} = \frac{5}{8}$.

The key is to notice that we don't care which numbers were picked, but only their odd/even nature.
Its answer would be like 
there will 16 cases such as 
(1,3),(1,5),(1,7),(1,9),(3,5),(3,7),(3,9),(5,7),(5,9),(7,9),(2,4),(2,6),(2,8),(4,6),(4,8),(6,8).
as given in question we want odd no. so there will be only 10 case(1st 10 cases).

10c1/16c1=10/16=5/8