Question: A box contains 3 white and 4 black balls. We draw a ball, if it is white we remove the obtained ball and add one black ball to the box, if it is black we remove the obtained ball and add one white ball to the box. After that we draw one more ball.
a) Given that the second ball is black, find the probability that the first ball was white.
b) Find the probability that the second ball is white.
Attempt:
a) The given that is being the second ball is black doesn't change the outcome of the first one. So it is 3/7.
b) I dont know how to approach this one. Should I try to solve it with cases. Case 1, first ball is white, then a white ball is taken out 2/7. Case 2, first ball is black, a white ball is added, 4/7. Then it is 6/7? Or if I am doing this in a conditional way, $P(w_2|w_1)=(\frac{P(w_1\cap w_2)}{P(w1)})$ Which doesn't makes sense to me because what is the intersection of that? The number of white balls? What am I missing out? Thanks a lot!
a)
Say x is the first pull and y is the second in the probability function P(x,y): $$P(x=W,y=B) = \frac{3}{7} \times \frac{5}{7}=\frac{15}{49}$$ $$P(x=B,y=B) = \frac{4}{7} \times \frac{3}{7}=\frac{12}{49}$$ $$P(x=W|y=B) = \frac{P(x=W,y=B)}{P(x=W,y=B)+P(x=B,y=B)}=\frac{5}{9}$$ You can't discount the condition you are given just because it occurs second.
b)
$$P(x=W,y=W) = \frac{3}{7} \times \frac{2}{7}=\frac{6}{49}$$ $$P(x=B,y=W) = \frac{4}{7} \times \frac{4}{7}=\frac{16}{49}$$ $$P(y=W) = P(x=W,y=W) + P(x=B,y=W) = \frac{22}{49}$$