It seems easy, but I don't get it

[Karlx]

Hi everybody.
Once more, I need your help.

The following is a problem that appears in Rohatgi's "An introduction to Probability and Statistics":

"Each of n urns contains four white and six black balls, while another urn contains five white and five black balls. An urn is chosen at random from the (n+1) urns, and two balls are drawn from it, both being black. The probability that five white and three black balls remain in the chosen urn is 1/7. Find n."

My first guess would be that n=6, but the solution given in the book is n=4.

Someone can tell me if I am wrong or there is a mistake in the book ?
Thanks.

 

[Karlx]

My guess is the following one:

The probability that five white and three black balls remain in the chosen urn is zero for the n urns that had initially only four white balls.
For the urn that had initially 5 white and 5 black balls, the probability of having 5 white and 3 black balls after having drawn 2 black balls is 1.
The probability that this former is the chosen urn is 1/(n+1).
So n must be equal to 6.

Am I wrong or there is a mistake in the book?
I would appreciate your hints.
Thanks

 

[awkward]

Let's say E is the event that the urn with 5 white and 5 balls is chosen, and B is the number of black balls drawn. You are told that

$$P(E \;|\; B=2) = 1/7$$.

By definition, $$P(E \;|\; B=2) = \frac{P(E \; \cap \; B=2)}{P(B=2)}$$.

Take it from there.

 

[Karlx]

Thanks for your answer, awkward.
This was my first attempt to the problem, but I wrote it in the following terms:

Let's say A is the event that five white and three black balls remain in the chosen urn, B is the number of black balls drawn and Cm is the event that the urn m has been chosen.
Then we are told that

$$P(A \;|\; B=2) = \frac{P(A \; \cap \; B=2)}{P(B=2)}$$

where
$$P(A \; \cap \; B=2) = \sum_{m=1}^{n+1}p(C_{m})p(A\; \cap \; B=2 \;|\; C_{m}) = \sum_{m=1}^{n+1} p(A\; \cap \; B=2 \cap \; C_{m})$$

and my problem was that I did not apply correctly the multiplication rule, that is

$$p(A\; \cap \; B=2 \cap \; C_{m}) = p(C_{m}) p(B=2 \;|\; C_{m}) p(A \;|\; B=2 \cap \; C_{m})$$

If I apply it correctly, then I obtain the result n=4, the same that appears in the book, and indeed the same if I apply your hint.
Thanks a lot.

 

[blue_raver22]

Hello! I understand that it can be frustrating when a problem appears to be easy, but you are having trouble understanding it. In this case, it is important to carefully review the information given and use logical reasoning to determine the correct solution. While your initial guess of n=6 may seem reasonable, the given solution of n=4 is actually correct. This is because the probability of selecting an urn with five white and three black balls, given that two black balls were drawn, is 1/7. This can only be true if the urn originally had six black balls, as stated in the problem. Therefore, the correct solution is n=4. It is always a good idea to double check your work and refer to reliable sources for clarification if needed. I hope this helps!