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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?

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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.

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  • $\begingroup$ thankyou! i know it is possible to do that way , but is there a way to find the probability function ? $\endgroup$ – fiksx May 21 at 13:58
  • $\begingroup$ @fiksx well, you already have now the complete probability distribution $\endgroup$ – Masacroso May 21 at 14:03
  • $\begingroup$ 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? $\endgroup$ – fiksx May 21 at 14:07
  • $\begingroup$ thankyou, i have never learn falling factorial , so its better to solve for every X? $\endgroup$ – fiksx May 21 at 14:37
  • $\begingroup$ @fiksx for this case it seems easier to solve directly one by one, for other cases maybe not $\endgroup$ – Masacroso May 21 at 15:58

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