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 $$P(b|A) = \frac{P(A|b)P(b)}{P(A)}$$ 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.
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.
I thought it would be instructive to try a different approach. 

If you drink three cups, that tells me for sure that two of the cups you drank (the first two) were not poisoned - otherwise you would not have gotten to the third cup.

So the poison must be in either cup 3 or cup 4, and since it is equally likely to be in any of the cups, there is a 50% chance of that happening.

Having survived the first two cups, you now have a 50-50 chance that cup 3 is poisoned (because it's either cup 3 or cup 4). To survive drinking cup 3, you need to beat those odds as well. 

Multiplying these probabilities, you once again get 0.25 for the chance of surviving.

Of course it's the same answer - but I thought this would give additional insight.