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Poisonous apples

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There are two bowls that you and a challenger must eat from.  After flipping a coin you were selected to pick the bowl that each would eat from.  In the first bowl there are three out of five poisonous apples.  In the second bowl, there are two out of five poisonous apples.  Whoever eats from the first bowl must eat two apples at random from the bowl.  Whoever eats from the second bowl must eat three random apples from the second bowl.  Which bowl should you pick to eat?

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8 answers to this question

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assuming five apples in each bowl,

Spoiler

you have a 2/5*1/4 chance to survive in the first bowl, or 10%. You have a 3/5*2/4*1/3 chance to survive with the second bowl, or a 10% chance. Its the same chance either way. I say it doesn't matter. (I assume that after eating an apple you take it out of the bowl.)

 

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Survival probabilities are (.4)^2 = .16 and (.6)^3 = .216 respectively for the two bowls. Pick bowl #2.

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the odds of eating a poison apple  for the 1st bowl  = 3/5 + 3/4 =  12/20 + 15/20 = 27/20 = !.35 chances of eating the P/A

                                                                          2nd bowl =2/5 + 2/4= 2/3 = 24 /60 +30/60 + 40/60 = 84/60 = 1.4 CHANCES 

Therefore you would eat from the 1st bowl with a 0.05 better chance of avoiding the P?A

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I agree with Flamebirde's answer. But I would add a bit more to it.

Spoiler

His answer would be correct if you assume that biting into a poisoned apple brings immediate death.

Pick from the first bowl and get
P(survival) = 2/5 x 1/4 = (2x1)/(5x4)
Pick from the second bowl and get
P(survival) = 3/5 x 2/4 x 1/3, but cancel out the 3 in the numerator and 3 in the denominator and you have (2x1)/(5x4) so the same thing.

However, if you don’t just keel over dead right after biting into a poison apple and just get a horrible stomachache or something, then it might be better if you only ate one poison apple instead of two, or two instead of three. If that’s the case (or in more general terms, if it’s better to eat fewer poisoned apples even if the probability of eating any non-zero number of poisoned apples is unchanged) then I’d rather eat from the bowl with only two poisoned apples. Since the question doesn’t specify whether apples are immediately lethal or not, that seems to me like the best course of action with the information that’s given.

Edit: Scratch that. If someone poisoned three of five apples and you had to choose two to eat, your probability of choosing one poisoned apple would be the same as if you chose three apples and then someone came by and randomly poisoned two of the five in play and you ask how many of those two apples that got poisoned are in your hands. It ends up being symmetrical, so it doesn't matter which bowl you pick.

 

Edited by plasmid
posted before thinking :p

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The OP does not say how many apples there are. It says the proportion that are poisonous.

Question: was that the intent?

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On 5/25/2018 at 6:25 AM, bonanova said:

The OP does not say how many apples there are. It says the proportion that are poisonous.

Question: was that the intent?

I meant them to be fractions so five in each

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Spoiler

Revised survival probabilities are

Bowl 1: (2/5)(1/4) = 0.1.
Bowl 2: (3/5)(2/4)(1/3) = 0.1.

Horrible odds. I would totally decline the challenge.

 

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Spoiler

I didn't realize probability was commutative that way. 2/5 choose three is equivalent odds to 3/5 choose two.

 

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