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

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

  • $\begingroup$ 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

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 $Y=1$, $X=4$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.