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There are $4$ boxes with $6$ white balls and $5$ black balls in each. You randomly select a ball from the first box. If the ball is white, the experiment is over. If it is black, you select a ball from the next box. The experiment is over if a white ball is drawn or if you draw a ball from all the boxes. Let $X$ be the number of boxes from which a ball was drawn. Let $Y$ be the number of white balls drawn.

Find the joint probability mass function?

I'm not asking for someone the solve the problem for me, I just need help how to start.

X 1 2 3 4
Y
0
1

How should I determine let's say, $P(X=1, Y=0)$? $P(X=1, Y=0) = P(X=1) * P(Y=0 | X=1) = 6/11 *$ ?

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  • $\begingroup$ I have updated the post to LaTeX, please see the editing is correct. $\endgroup$ – Jeel Shah Aug 19 '13 at 20:06
  • $\begingroup$ It seems you have to think more about "the physics" of the experiment before embarking in symbols manipulations. What does it mean, say, that X=3? And why, if X=3, Y can take exactly one value? $\endgroup$ – Did Aug 19 '13 at 20:11
  • $\begingroup$ ^agreed. A good place to start would be to determine the pmf of X and Y individually. For example, for Y=0, the only way this occurs is if a black ball is chosen from each of the 4 boxes, which translates into (5/11)^4. $\endgroup$ – user79790 Aug 19 '13 at 20:15
  • $\begingroup$ I have yet to understand how one can pass from a state of complete puzzlement about a question to a state of complete understanding in less than 8 minutes. Well... $\endgroup$ – Did Aug 19 '13 at 20:30
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First the easy part. The probability that $Y=0$ is $\left(\frac{5}{11}\right)^4$. For we must get $4$ black in a row.

In that case $X=4$. So in the $Y=0$ row of your table, we will have the entries $0\quad 0\quad 0 \quad \left(\frac{5}{11}\right)^4$.

Now we fill in the $Y=1$ row. The probability that $Y=1$ and $X=1$ is $\frac{6}{11}$. For $Y=1$, $X=1$ happens precisely if we draw a white from the first box.

The probability that $Y=1$ and $X=2$ is not much harder. We must draw a black from the first, then a white from the second. The probability is $\frac{5}{11}\cdot\frac{6}{11}$.

I leave it to you to fill in the entries for $Y=1$, $X=3$ and $Y=1$, $X=4$.

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