- #1
Gauss M.D.
- 153
- 1
Homework Statement
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?
Homework Equations
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?