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This is a problem that I saw in a previous exam in my university:

There are 3 types of balls in a box: white, blue and red (their quantity is unknown). We pull out a ball from a box randomly, and if it is not red- we return it to the box. We keep doing it until we get a red ball, and then we stop pulling out balls. Let $X$ be the number of blue balls pulled out, $Y$ the number of white balls pulled out and $L$ the number of total balls that were pulled out. Find the distribution of $X|L=k$ ,$Y|L=k$ ,$L$.

My attempt: it’s easy to tell that $(X|L=k), (Y|L=k)$~$B(k-1,p_1)$ and $L$~$G(p_2)$. Now we need to find $p_1$ and $p_2$. $p_1$ is the probability to pull out blue ball (white ball for $(Y|L=k)$, It’s symmetric) and $p_2$ is the probability to pull out red ball. The problem is that the quantity of the balls is unknown, and therefore I don’t know how to calculate $p_1,p_2$. I saw in the solution of that exam that $p_1=0.5 , p_2=1/3$ and I don’t understand why. If for example we have one blue ball, 10 white balls and 10 red balls in the box- then $p_1,p_2$ are not correct.

Can you explain why $p_1=0.5 , p_2=1/3$ ?

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My attempt: it’s easy to tell that $(X|L=k),(Y|L=k)\sim B(k−1,p_1)$ and $L\sim G(p_2)$.

To be sure, $(X\mid L{=}k)\sim\mathcal {Bin}(k-1,p_1)~,~ (Y\mid L{=}k)\sim\mathcal{Bin}(k-1,1-p_1)~,~ L\sim\mathcal{Geo}_1(p_2)$

Where if the number of red, blue, and white balls are $r,b,w$ we have $p_1=\frac{b}{b+w}, p_2=\frac r{r+w+b}$

Then $p_1=1/2, p_2=1/3$ happens when we have $r=b=w$; that is the same amount for each colour of ball.

Either there is some information missing from the problem, or you are expected to just assume that there are an equal quantity of balls of each type in the box as a Bayesian Prior.

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    why p1=b/(b+w) ? isn't it suppose to be b/(b+w+r)?2017-02-14
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    @john $X\mid L{=}k$ is the count of black balls among the $k-1$ balls that are not the first red ball (when given that happens on draw $k$). $~p_1$ is the conditional probability of selecting a black ball when given that it is not red.2017-02-14