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 $P(I_i=1) = 9/12$,$P(I_2=1)=8/11$..
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?