I have tried to find a solution for this problem a lot but I couldn't solve it.
Consider a box containing 3 red ball and 7 white balls. Suppose that balls are drawn one at a time, at random, without replacement from this box until two red balls are obtained. Let X denote the number of the draw on which the second red ball is obtained. Find the probability mass function of X
all what I could do is to get the range for X. where X = 2 , 3, 4,5,6,7,8,9
Let's compute $P(X=i)$.
To be in that scenario, the first $i-1$ choices must consist of exactly $1$ red and $i-2$ whites. There are $\binom 7{i-2}$ ways to choose the whites and $3$ ways to choose the reds. Thus there are $3\times \binom 7{i-2}$ ways to choose the first $i-1$. Of course there are $\binom {10}{i-1}$ ways to do it with out restriction. Thus the probability that the first $i-1$ contain exactly $1$ red is $$3\times \binom 7{i-2}\Big /\binom {10}{i-1}$$
Having successfully chosen the first $i-1$ we now require that the $i^{th}$ choice be red. That has probability $\frac 2{10-(i-1)}=\frac 2{11-i}$. Thus the answer is $$\left(3\times \binom 7{i-2}\Big /\binom {10}{i-1}\right)\times \frac 2{11-i}$$
