We have two boxes:
$A$ - with two 8 balls, and $B$ - with 4 balls.
we choose randomly box and pull out a ball.
We do it again and again until box $B$ will be empty.
What is the probability that box $A$ is empty?
(When we choose $B$ and see it empty).
I'm stuck and I don't have any idea how to begin to solve this problem....
I know that I need to calculate $P(E|F)$, where $E$ = Box $A$ is empty, $F$ = Box $B$ is empty.
Thank you!
On
I think I understand:
It's a Bernoulli Experiment:
$$X\sim Bin\left(12,\frac12\right)$$
When $X$= "the box we choose".
We have to caluculate: $P(X=4)$ (At 12 times that we make this experiment we have to choose box $B$ 4 times ("success") and box $A$ 8 times ("failed")).
So the result is:
$$P(X=4)=\binom{12}{4}\cdot \left(\frac12\right)^4\cdot \left(\frac12\right)^8=\frac{495}{4096}$$
On
$A$ will be empty if it has been picked eight or more times in the first 11 rounds.
Probability of picking $A$ exactly 8 times out of 11: $\binom{11}{8}2^{-11}$
Probability of picking $A$ exactly 9 times out of 11: $\binom{11}{9}2^{-11}$
Probability of picking $A$ exactly 10 times out of 11: $\binom{11}{10}2^{-11}$
Probability of picking $A$ exactly 11 times out of 11: $\binom{11}{11}2^{-11}$
Adding these up: $$(165 + 55 + 11 + 1)2^{-11} = \frac{29}{256}$$
When B is empty, the possibilities are you have done from four "picks" (you hit the B every time) to 8 + 4 = 12 "picks" (you hit A at least 8 times before taking the last ball from B, for which you must have hit B three times before). You could not have done more or less "picks".
So there are 9 possibilities (4, 5, 6, 7, 8, 9, 10, 11, 12 "picks"), and the question is: "what is the probability that you will have done 12 "picks" (which is the only outcome in which A is empty)?" The answer seems simple, the probability for one out of nine possible outcomes is 1/9, right?
Right?
(This answer is purposefully incomplete, It's meant to give you a start on the solution but not to give you the complete answer. So now, try to solve the problem considering what I've written, and if you still can't figure it out, I'll post the whole solution in a few hours).
Addition
When B has been emptied, there are 9 possible "amounts of picks" that could have got you here. But all these are not equally likely, for example: there is only one way to empty B in four picks, namely by picking {B, B, B ,B}. But there are several ways of emptying B in five picks, to name two: {A, B, B, B, B}, {B, A, B, B, B}. So to calculate the likelihood of emptying B in 12 picks, one would have to find out how many "ways" that one can be done, versus the summarized "ways" of all the other 8 amounts of picks. One way to find this out would be by hand, but perhaps you can fins a pattern or formula for computing "how many ways can you order (for instance) five B's and two A's?"
Addition II
Well, see Graham Kemp's comment for the rest of the solution, I think he did a great job.