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?

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.

NicNic8 has provided a nice intuitive answer to the question.

Here are three alternative methods. In the first, we solve the problem directly by considering which cups are selected if she is poisoned. In the second, we solve the problem indirectly by considering the order in which the cups are selected if she is not poisoned. In the third, we add the probabilities that she was poisoned with the first cup, second cup, or third cup.

Method 1: We use the hypergeometric distribution.

There are $\binom{4}{3}$ ways to select three of the four cups. Of these, the person selecting the cups is poisoned if she selects the poisoned cup and two of the three cups of water, which can be done in $\binom{1}{1}\binom{3}{2}$ ways. Hence, the probability that she is poisoned is $$\Pr(\text{poisoned}) = \frac{\binom{1}{1}\binom{3}{2}}{\binom{4}{3}} = \frac{1 \cdot 3}{4} = \frac{3}{4}$$

Method 2: We subtract the probability that she is not poisoned from $1$.

The probability that the first cup she selects is not poisoned is $3/4$ since three of the four cups do not contain poison. If the first cup she selects is not poisoned, the probability that the second cup she selects is not poisoned is $2/3$ since two of the three remaining cups do not contain poison. If both of the first two cups she selects are not poisoned, the probability that the third cup she selects is also not poisoned is $1/2$ since one of the two remaining cups is not poisoned. Hence, the probability that she is not poisoned if she drinks three of the four cups is $$\Pr(\text{not poisoned}) = \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{4}$$ Hence, the probability that she is poisoned is $$\Pr(\text{poisoned}) = 1 - \Pr(\text{not poisoned}) = 1 - \frac{1}{4} = \frac{3}{4}$$

Addendum: We can relate this method to the first method by using the hypergeometric distribution.

She is not poisoned if she selects all three cups which do not contain poison when selecting three of the four cups. Hence, the probability that she is not poisoned is $$\Pr(\text{not poisoned}) = \frac{\dbinom{3}{3}}{\dbinom{4}{3}} = \frac{1}{4}$$ so the probability she is poisoned is $$\Pr(\text{poisoned}) = 1 - \frac{\dbinom{3}{3}}{\dbinom{4}{3}} = 1 - \frac{1}{4} = \frac{3}{4}$$

Method 3: We calculate the probability that the person is poisoned by adding the probabilities that she is poisoned with the first cup, the second cup, and the third cup.

Let $P_k$ denote the event that she is poisoned with the $k$th cup.

Since there are four cups, of which just one contains poison, the probability that she is poisoned with her first cup is $$\Pr(P_1) = \frac{1}{4}$$

To be poisoned with the second cup, she must not have been poisoned with the first cup and then be poisoned with the second cup. The probability that she is not poisoned with the first cup is $\Pr(P_1^C) = 1 - 1/4 = 3/4$. If she is not poisoned with the first cup, there are three cups remaining of which one is poisoned, so the probability that she is poisoned with the second cup if she is not poisoned with the first is $\Pr(P_2 \mid P_1^C) = 1/3$. Hence, the probability that she is poisoned with the second cup is $$\Pr(P_2) = \Pr(P_2 \mid P_1^C)\Pr(P_1) = \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$$

To be poisoned with the third cup, she must not have been poisoned with the first two cups and then be poisoned with the third cup. The probability that she is not poisoned with the first cup is $\Pr(P_1^C) = 3/4$. The probability that she is not poisoned with the second cup given that she was not poisoned with the first is $\Pr(P_2^C \mid P_1^C) = 1 - \Pr(P_2 \mid P_1^C) = 1 - 1/3 = 2/3$. If neither of the first two cups she drank was poisoned, two cups are left, one of which is poisoned, so the probability that the third cup she drinks is poisoned given that the first two were not is $\Pr(P_3 \mid P_1^C \cap P_2^C) = 1/2$. Hence, the probability that she is poisoned with the third cup is $$\Pr(P_3) = \Pr(P_3 \mid P_1^C \cap P_2^C)\Pr(P_2^C \mid P_1^C)\Pr(P_1^C) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}$$ Hence, the probability that she is poisoned is $$\Pr(\text{poisoned}) = \Pr(P_1) + \Pr(P_2) + \Pr(P_3) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$$

Not sure why everybody uses such a complicated approach:
after drinking three cups, one remains. The chance that she is alive is equal to the chance that the remaining cup is the poison, which is one in four = 25%.

The sequence of drinking water and or poison is completely irrelevant.