1. May 22, 2013

### Gauss M.D.

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$\cap$B) = (3c2*7c2*1c1*5c5 + 3c1*7c3*2c1*6c4 + 3c1*7c3*2c2*6c3)/(10c4*6c5)

P(B|A) = P(A$\cap$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?

2. May 22, 2013

### HallsofIvy

Either of Peter or Simon will be poisoned if they eat at least one poisoned fruit, right? So it is sufficient to calculate the probabllity they get NO poisoned fruit, then subtract from 1.

3. May 22, 2013

### verty

This is not correct.

4. May 22, 2013

### haruspex

It asks for the prob that both get poisoned, not that at least one gets poisoned. If you're applying that individually it's true, but that still doesn't deal with the joint probability.
Or maybe you meant this: P(A&B) = 1 - P(!A) - P(!B) + P(!A&!B) (! signifying NOT).

5. May 23, 2013

### verty

Hint: Get a formula for P(A or B).