Bag contains 9 blue balls, 3 red balls. A ball is selected at random, colour recorded and NOT replaced. A second ball is selected at random, colour recorded. a) Draw tree diagram (no problem). Part 2 a) Find probability second ball is red (no problem, 0.25) b) Find probability both balls selected are red, given that the second ball is red. I'm having a problem understanding where my logic is wrong. Using the formula: P(B|A) = P(B and A) / P(A) or P(both balls are red | second ball is red) = P(both balls are red and second ball is red) / P(second ball is red) P(second ball is red) = (1/4) or 0.25 (from part a) P(both balls are red) = (3/12) * (2/11) = (1/22) Because I need to find: P(both balls are red and second ball is red) I do the following to get the intersection (1/4) * (1/22) = (1/88) and then divide it by P(second balls is read) to reduce the sample space down to the given event. However this leads to the wrong answer. Looking at the mark scheme, the just do (3/12) * (2/11) to get the intersection, but following the formula leads me to a different answer. Looking at the textbook it says that if two events are independent then P(A|B) = P(A), if so then these two events are not independent, and the mark scheme is treating it as if they were independent.