Jump to content
BrainDen.com - Brain Teasers
  • 0
BMAD

Poisonous apples

Question

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?

Share this post


Link to post
Share on other sites

10 answers to this question

  • 0

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

Share this post


Link to post
Share on other sites
  • 0

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.)

 

Share this post


Link to post
Share on other sites
  • 0

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

Share this post


Link to post
Share on other sites
  • 0

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

Share this post


Link to post
Share on other sites
  • 0

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

Question: was that the intent?

Share this post


Link to post
Share on other sites
  • 0
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

Share this post


Link to post
Share on other sites
  • 0
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.

 

Share this post


Link to post
Share on other sites
  • 0
Spoiler

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

 

Share this post


Link to post
Share on other sites
  • 1

Probabilities of something occuring in sequence multiply together, they don't add.

We could calculate the probability where we don't eat 2 poisonous or where we do.  I chose to calculate where we do.  

For the first bowl.  We only care if you eat 2 poisonous apples. The probability the first apple eaten is poisonous is 3/5.   Assuming you ate a poisonous apple because that is the only scenario we care about, the probability the second apple is poisonous is 2/4.  This is the only scenario eating from this bowl that we die.  Multiply this together (3/5 * 2/4) and the chance of dying is 30%.   70% chance you survive.

For the second bowl, we do the same thing, but there are multiple possibility branches for this one.

The probability the first apple eaten is poisonous is 2/5.  The probability the second apple is poisonous is 1/4 if you already ate a poisonous one.
So a 10% chance you die after eating the second apple (assuming it is fast poison).  If you are still alive then the 2nd apple was not poisonous so that had a probability of 3/4.  That makes he probability for the third equal to 1/3. So 2/5 * 3/4 * 1/3 is 10% chance of dying.  
The chance the first apple is not poisonous is 3/5.  So for the second the chance is 2/4, leaving the last at 1/3.    So multiply these out and you get 10%%.  
Due to having 3 scenarios that result in death we have to add those probabilities.  10% + 10% + 10% = 30%.  70% chance we survive.

So both bowls have the same odds.  Others arrived at this answer, but with the wrong math.  Cheers.

Edited by Taciav

Share this post


Link to post
Share on other sites
  • 0

OP forgot to state that eating 2 poison apples results in death, while eating only 1 will not cause harm.

Therefore Taciav has the correct solution.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

  • Recently Browsing   0 members

    No registered users viewing this page.

×