# 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 $\alpha, \beta, \gamma, \delta$ with events

• $A = \{\alpha \text{ is water}\}, \ a = \{\alpha \text{ is poison}\}$
• $B = \{\beta \text{ is water}\},\ b = \{\beta \text{ is poison}\}$
• $C = \{\gamma \text{ is water}\},\ c = \{\gamma \text{ is poison}\}$
• $D = \{\delta \text{ is water}\},\ d = \{\delta \text{ is poison}\}$

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

My doubt is that I shouldn’t order the cups because that assumes $\delta$ 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.