What Is the Probability That Both Peter and Simon Get Poisoned?

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Homework Help Overview

The problem involves calculating the probability that both Peter and Simon get poisoned from a bag of fruits, given certain conditions about their consumption and the dog's choice of fruit. The context is rooted in probability theory, specifically dealing with conditional probabilities and combinatorial selections.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss reformulating the problem based on the dog's consumption of a healthy fruit and explore different probability calculations for both Peter and Simon getting poisoned. Some question the correctness of the probability formulas being used.

Discussion Status

The discussion is ongoing, with participants offering different perspectives on how to approach the problem. Some have suggested alternative methods to calculate the probabilities, while others are questioning the validity of the current approaches being discussed.

Contextual Notes

There is a focus on the joint probability of both individuals getting poisoned, with some participants noting the distinction between calculating the probability of at least one being poisoned versus both. The complexity of the problem is acknowledged, with references to the combinatorial nature of the calculations involved.

Gauss M.D.
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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\capB) = (3c2*7c2*1c1*5c5 + 3c1*7c3*2c1*6c4 + 3c1*7c3*2c2*6c3)/(10c4*6c5)

P(B|A) = P(A\capB)/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?
 
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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.
 
Gauss M.D. said:
P(B|A) = P(A\capB)/P(B)

This is not correct.
 
HallsofIvy said:
Either of Peter or Simon will be poisoned if they eat at least one poisoned fruit, right? So it is sufficient to calculate the probability they get NO poisoned fruit, then subtract from 1.
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).
 
Hint: Get a formula for P(A or B).