1. The problem statement, all variables and given/known data Peter and Simon shares a bag of 11 fruits, of which 3 are poisoned. Peter eats 4 fruits, Simon eats 3 and their dog eats 1 fruit. What is the probability that both Peter and Simon gets poisoned, given that the dog ate a healthy fruit? 2. Relevant equations 3. The attempt at a solution The dog simply removed a healthy fruit, so we can reformulate the question as: Given 10 fruits of which 3 are poisoned, what is the probability both Peter (4 fruits) and Simon (6 fruits) gets poisoned? A = Peter gets poisoned B = Simon gets poisoned XcY = X choose Y P(A) = (3c3*7c1 + 3c2*7c2 + 3c1*7c3)/10c4 P(A[itex]\cap[/itex]B) = (3c2*7c2*1c1*5c5 + 3c1*7c3*2c1*6c4 + 3c1*7c3*2c2*6c3)/(10c4*6c5) P(B|A) = P(A[itex]\cap[/itex]B)/P(B) This should work I believe. But I feel like I'm deriving the Fourier heat equation to figure out the boiling point of water. Is there an easier way?