Shouldn't it be 1-(5c2/10c2) =1-(10/45)=7/9 ?

We want P(select at least 1 even numbered ball)
When it comes to probability questions involving "at least," it's usually best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(getting at least 1 even) = 1 - P(not getting at least 1 even)
What does it mean to not get at least 1 even? It means getting zero evens.
So, we can write: P(getting at least 1 even) = 1 - P(getting zero evens)

P(getting zero evens)
P(getting zero evens) = P(getting both odds)
= P(1st ball is odd AND 2nd ball is odd)
= P(1st ball is odd) x P(2nd ball is odd)
= 1/2 x 1/2
= 1/4

Aside: Since we are replacing the ball after the first selection, the probability of getting an odd number to ball is 5/10 (aka 1/2) with each selection

So...P(select at least 1 even numbered ball) = 1 - 1/4 = 3/4

Answer: C

Cheers,
Brent