Probability, $2$ urns, one with $2$ black, and one with a white and a black ball.

We have $2$ opaque bags, each containing $2$ balls. One bag has $2$ black balls and the other has a black ball and a white ball. You pick a bag at random and then pick one of the balls in that bag at random. When you look at the ball, it is black. You now pick the second ball from that same bag. What is the probability that this ball is also black?

I am constantly getting this question's answer wrong whenever I attempt it. So please tell me what I am doing wrong here ?

My Solution:

Let $U_1$ be the urn with two black balls and $U_2$ be the urn with one black and one white ball.

$P(U_1)$ : Probability of selecting $U_1 = 1/2$

$P(U_2)$ : Probability of selecting $U_2 = 1/2$

$P_1(B|B)$ : Probability of second draw is black, given the first draw from the same urn, $U_1$, is black = $1$

$P_2(B|B)$ : Probability of second draw is black, given the first draw from the same urn, $U_2$, is black = $0$

Then we have,

$P(U_1) * P_1(B|B) + P(U_2) * P_2(B|B) = 1/2$

But the answer is $2/3$, so what am I missing here ?

• With regard to your question, you are asked for the conditional probability that second ball is black given that the first ball is black, but you are calculating an unconditional probability. What you are missing is that the first black ball is equally likely to be any of the three black balls, and in two of those three cases, the other ball is black while in one of the three cases, the other ball is white. – Dilip Sarwate Jan 15 '13 at 18:15

So, if you have drawn out one of the three black balls, the probabilities that it came from urn $U_i$ are
$$P(U_1|B) = 2/3$$ $$P(U_2|B) = 1/3$$
If it's hard to understand, you can think of a similar situation - let's say you have an $U_1'$ with 100 black balls and $U_2'$ with 99 white balls and one black ball. If you take one urn and pull out a ball and it turns out to be black, you can be quite sure that you got $U_1'$. In this case it's the same, just not that obvious.