gmatravi wrote:

Bunuel,

I think I am missing something here. I tried with an example:

Lets blue balls be 3 and red balls be 4 :

4/7 * 3/6 = 2/7

but if

B=5 and R=6 then it will be 3/11

How you are getting 1/2 as the answer?

What you've calculated here is conditional probability.

3/11 would've been correct if we were asked to calculate the probability of getting the

FIRST ball red and

THEN picking the second ball blue.

But here we're given that the

FIRST has already been picked and it was red. That event has already happened, so we don't need to worry about that.

In mathematical terms,

P[first ball red] = 1 , as there is no other possible outcome other than what already came

Using your example,

Let blue balls be 3 and red balls be 4 :

1/1 * 3/6 = 1/2

Hence, it's correct.