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 ?
 A: You are missing the information that you've got from the fact that the first ball is black.
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.
A: 
If we think of throwing all the balls in a big pool, Then that pool will have BIG POOL = (B1,B1,B2,W2) = 3 BLACK BALLS of which 2 are from U1. According to the problem, we shall select this U1.
Another way :
lets say :  U1 = 2B and U2 : 1W + 1B
$P(U_1|B)=\dfrac{P(B|U_1)\cdot P(U_1)}{P(B|U_1)\cdot P(U_1)+P(B|U_2)\cdot P(U_2)}=\dfrac{1\cdot\dfrac12}{1\cdot\dfrac12+\dfrac12\cdot\dfrac12}=\dfrac23$
$P(U_2|B)=\dfrac13$
The probability that the BLACK ball is drawn the Second time is
$\dfrac23\cdot1+\dfrac13\cdot(0) =\dfrac{2}{3}$
