# drawing 2 red balls from box contains 3 red balls and 7 white balls

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

• Where are you getting stuck? Just go one by one. Easy to see that $P(X=2)=\frac 3{10}\times \frac 2{9}$ for example. Now do the rest. – lulu Jan 20 '19 at 17:14
• the problem here is i can't know how many times i will draw balls so i can not decide the probability very well. – Mo'men Mustafa Jan 20 '19 at 17:22
• maybe i will draw one red ball then 3 white balls then another red ball i won't draw two red balls after each other – Mo'men Mustafa Jan 20 '19 at 17:23
• It's really not that hard. Try some of the cases and I think you'll get the idea. In any case, people here will meet you half way if you show a little effort. What's $P(X=3)$ for instance? – lulu Jan 20 '19 at 17:33
• okay I think it will be (((3C1)*(7C1))/(10C2))*2/8 – Mo'men Mustafa Jan 20 '19 at 17:58

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 $$\left(3\times \binom 7{i-2}\Big /\binom {10}{i-1}\right)\times \frac 2{11-i}$$