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?

 To determine the probability that the second ball is also black, given that the first ball you drew was black, we can use conditional probability.

Let's define the events: - B1: The first ball drawn is black. - B2: The second ball drawn (from the same bag) is black. - M: The bag chosen is the one with two black balls. - N: The bag chosen is the one with one black and one white ball.

We are looking for the probability of B2 given that B1 has occurred, which is noted as P(B2|B1).

Now, consider P(B1). This is the probability that you draw a black ball in the first draw and it can come from either of the two bags. If you pick the all-black ball bag (M), the probability of drawing a black ball is 1. If you pick the mixed ball bag (N), the probability of drawing a black ball is 1/2. Given that the total probability of picking either bag is 1/2, you can calculate P(B1) as follows:

P(B1) = P(M) * P(B1|M) + P(N) * P(B1|N) P(B1) = 1/2 * 1 + 1/2 * 1/2 P(B1) = 1/2 + 1/4 P(B1) = 3/4 or 0.75

Now, let's compute P(B2|B1). To find this, we only consider scenarios where B1 has occurred (the first ball is black). The only way B2 can occur is if we originally picked the all-black ball bag (M):

P(B2|B1) = P(M|B1) * P(B2|M)

We know P(B2|M) = 1, since if you've picked the all-black ball bag, the second ball you draw will definitely be black. Now we need P(M|B1), which is the probability that we picked the all-black ball bag given that we drew a black ball.

By Bayes' theorem, we can find P(M|B1) as:

P(M|B1) = P(B1|M) * P(M) / P(B1) P(M|B1) = 1 * 1/2 / (3/4) P(M|B1) = 1/2 / (3/4) P(M|B1) = 2/3

Now we substitute this back into our formula for P(B2|B1):

P(B2|B1) = P(M|B1) * P(B2|M) P(B2|B1) = (2/3) * 1 P(B2|B1) = 2/3

So the probability that the second ball is also black, given that the first ball drawn was black, is 2/3 or approximately 0.6667.