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

  • $\begingroup$ 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. $\endgroup$ – lulu Jan 20 '19 at 17:14
  • $\begingroup$ the problem here is i can't know how many times i will draw balls so i can not decide the probability very well. $\endgroup$ – Mo'men Mustafa Jan 20 '19 at 17:22
  • $\begingroup$ 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 $\endgroup$ – Mo'men Mustafa Jan 20 '19 at 17:23
  • $\begingroup$ 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? $\endgroup$ – lulu Jan 20 '19 at 17:33
  • $\begingroup$ okay I think it will be (((3C1)*(7C1))/(10C2))*2/8 $\endgroup$ – 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 $$\left(3\times \binom 7{i-2}\Big /\binom {10}{i-1}\right)\times \frac 2{11-i}$$

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