There are 4 cups of liquid. Three are water and one is poison. If you were to drink 3 of the 4 cups, what is the probability of being poisoned?

In Season 5 Episode 16 of Agents of Shield, one of the characters decides to prove she can’t die by pouring three glasses of water and one of poison; she then randomly drinks three of the four cups. I was wondering how to compute the probability of her drinking the one with poison.

I thought to label the four cups α,β,γ,δ with events

  • A={α is water}, a={α is poison}
  • B={β is water}, b={β is poison}
  • C={γ is water}, c={γ is poison}
  • D={δ is water}, d={δ is poison}

If she were to drink in order, then I would calculate P(a)=1/4.
Next P(b|A)=P(A|b)P(b)P(A) Next P(c|AB), which I’m not completely sure how to calculate.

My doubt is that I shouldn’t order the cups because that assumes δ is the poisoned cup. I am also unsure how I would calculate the conditional probabilities (I know about Bayes theorem, I mean more what numbers to put in the particular case). Thank you for you help.


The probability of not being poisoned is exactly the same as the following problem:

You choose one cup and drink from the other three. What is the probability of choosing the poisoned cup (and not being poisoned)? That probability is 1/4.

Therefore, the probability of being poisoned is 3/4.

Source : Link , Question Author : Eli , Answer Author : NicNic8

Leave a Comment