# Finding probability distribution of $X$

In the box, there are 3 red balls and 3 blue balls, and from this box, the extraction of the ball is continued until the blue ball comes out. $$X$$ denotes extracted balls until blue ball comes out. Find the probability distribution, and find the mean and variance of $$X$$.

I am trying to find the function for probability distribution $$\frac{\binom{3}{x-1} \times \binom{3}{1}}{\binom{6}{x-1}}$$ , but I don't get the right answer. Which part is wrong?

Note that $$\Pr[X=1]=\frac36$$ because you get a blue ball from a group of $$3$$ blue balls from a total number of $$6$$ balls in the box.

Now $$\Pr[X=2]=\frac36\cdot\frac35$$ because you get first one red ball out of a group of $$3$$ red balls from a total of $$6$$ balls and after you took a blue ball from the group of $$3$$ blue balls from a total of $$5$$ balls in the box.

Continuing this reasoning we find that $$\Pr[X=3]=\frac36\cdot\frac25\cdot\frac34$$ and $$\Pr[X=4]=\frac36\cdot\frac25\cdot\frac14\cdot\frac 33$$. By last note that for $$k>4$$ we have that $$\Pr[X=k]=0$$ by the pigeonhole principle.

The probability mass function can be written compactly as $$f_X(k)=\frac{3^{\underline{k-1}}\cdot3}{6^\underline k}$$ where $$n^\underline k$$ is a falling factorial.

• thankyou! i know it is possible to do that way , but is there a way to find the probability function ? – fiksx May 21 at 13:58
• @fiksx well, you already have now the complete probability distribution – Masacroso May 21 at 14:03
• i was thinking , $\frac{\binom{3}{x-1} \times \binom{3}{1}}{\binom{6}{x-1}}$ , choosing red for $x-1$ and 3 blue , but it is seems not right? – fiksx May 21 at 14:07
• thankyou, i have never learn falling factorial , so its better to solve for every X? – fiksx May 21 at 14:37
• @fiksx for this case it seems easier to solve directly one by one, for other cases maybe not – Masacroso May 21 at 15:58