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There are 3 red, 9 green balls in the box. We pick them randomly without returning. When take out red ball for the first time we stop. X-discrete distributed variable, where we have the number of taken balls before we take out first red ball.

What is the probability distribution for X?

So we can write $X$ is the following way:

$X= I_1+I_2+...+I_9$, where $I_i = 1$ if we take out green ball and $I_i=0$ otherwise. Hence $P(X) = P(I_1+...+I_9) = P(I_1=1)+ ... + P(I_9 =1)$. Now the probability green ball is drawn is $9/12$ hence $P(I_i=1) = 9/12$,$P(I_2=1)=8/11$..

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  • $\begingroup$ Why $9$ entries? At least one of the first $8$ draws is certain to be red. $\endgroup$
    – lulu
    Commented Aug 26, 2018 at 12:47
  • $\begingroup$ I made an edit. $\endgroup$
    – user557550
    Commented Aug 26, 2018 at 12:48
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    $\begingroup$ It's still wrong. Now all you know is that one of the first $10$ draws must be red, so the distribution contains $10$ entries. Nor do I understand your computation. $X=1$ means the first draw is red, so $\frac 3{12}$. $\,\,$ $X=2$ means the first draw is green and the second red, so $\frac 9{12}\times \frac 3{11}$. And so on. $\endgroup$
    – lulu
    Commented Aug 26, 2018 at 12:49
  • $\begingroup$ "So first time we will have $\frac9{12}$ at first.." What exactly do you mean by that? Also note that $X$ takes values in $\{0,1,2,\dots,9\}$ so can take $10$ values as @lulu remarks. Repair and make things more clear in the question you pose by means of an edit (not in a comment). $\endgroup$
    – drhab
    Commented Aug 26, 2018 at 13:05
  • $\begingroup$ For clarity: the computations I did in my context incorrectly assumed that $X$ was the number of the first red draw (instead of the number of green draws which precede it). Thus where I wrote $X=1$ I should have written $X=0$, and so on. $\endgroup$
    – lulu
    Commented Aug 26, 2018 at 13:13

1 Answer 1

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Let $G_i$ denote the event that the $i$-th ball drawn is green.

Let $R_i$ denote the event that the $i$-th ball drawn is red.

The event that no green balls are drawn when we take out the first red ball is the same as the event that the first ball drawn is a red ball. So we find:$$P(X=0)=P(R_1)=\frac3{12}$$

The event that $1$ ball is drawn when we take out the first red ball is the same as the event that the first ball drawn is a green ball and the second is a red ball. So we find:$$P(X=1)=P(G_1\cap R_2)=P(G_1)P(R_2\mid G_1)=\frac9{12}\frac3{11} $$

The event that $2$ balls are drawn when we take out the first red ball is the same as the event that the first $2$ balls drawn are green and the third is red. So we find:$$P(X=2)=P(G_1\cap G_2\cap R_3)=P(G_1\cap G_2)P(R_3\mid G_1\cap G_2)=$$$$P(G_1)P(G_2\mid G_1)P(R_3\mid G_1\cap G_2)=\frac9{12}\frac8{11}\frac3{10} $$

et cetera.

Can you take it from here?

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  • $\begingroup$ So we continue until $P(X=9)$, and we have that $P(X=3)= 9/12*8/11*7/10*3/9$, $P(X=4)= 9/12*8/11*7/10*6/9*3/8$, and so on and then $P(X=9)= 9/12*8/11*...*1/4*3/3$. And then we need to add all of those? $P(X)= P(X=0)+p(X=1)+...+P(X=9)$? $\endgroup$
    – user557550
    Commented Aug 26, 2018 at 14:10
  • $\begingroup$ You describe the correct way for finding $P(X=i)$ for $i=0,1,\dots,9$. If you have done that then you are ready: you found the probability distribution of $X$. You can check yourself on mistakes. If you did not make any then the summation of the found probabilities is $1$. But that is not a need. Again the equality in your comment makes no sense because the expression $P(X)$ on LHS makes no sense. Arguments of $P$ are events and $X$ is no event. It is a random variable. $\endgroup$
    – drhab
    Commented Aug 26, 2018 at 14:47

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