# It seems easy, but I don't get it

1. Aug 27, 2010

### 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.

2. Aug 28, 2010

### 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?
Thanks

3. Aug 28, 2010

### 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.

4. Aug 30, 2010

### Karlx

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.

Last edited: Aug 30, 2010