the question is :
(b) Box 1 contains six red balls and four green balls and box 2 contains seven red balls and three green balls. A ball is chosen at random from box 1 and placed in box 2. A ball is then chosen at random from box 2 and placed in box 1.
(i) the probability that a red ball is selected from both box 1 and box 2;
(ii) the probability that, at the conclusion of the selection process, there are again six red balls and four green balls in box 1.
a) Pr( red in the first time and red in the second) = Pr(red in the first). Pr(red in the second | red in the first) (conditional probability)
= 6/10 . 8/11 = 24/55
b) A: the event red in the first and red in the second moving .
Ag: the event green in the first moving and green in the second
B: the event that after 2 moving there are again 6 red and 4 green in the box 1. We need calculate Pr(A|B) = ?
Pr(A|B) = Pr(AB)/ Pr(B) = Pr(A)/Pr(B) (because A \subset B)
We calculate Pr(B):
Pr(B) = Pr(A) + Pr(Ag) --> the result