Probability Question: Bayes' theorem, Binomial Coefficient

Question

From a group of 20 people (including Alice and Bob), 5 are randomly selected. Assuming that Alice was selected, what is the probability that Bob will not be selected?

This is the question I faced in my probability course.

My solution was to calculate all possible selections when Alice is selected: ${19 \choose 4}$, when both Alice and Bob are selected ${18 \choose 3}$.

Then, I calculated the probability for Bob not to be selected if Alice is selected:

$$1-{\frac{18 \choose 3}{19 \choose 4}} = {\frac{15}{19}} \approx 0.78947$$

I would be happy if someone will confirm that this solution is correct, or tell me what is the correct solution.

Answer 1

Your result is correct.

Another way is to say that, if Alice is selected, there are $4$ places left and $19$ people so $15$ will not be selected, and each of the $19$ is equally likely to not be selected, making the probability of Bob not being selected $\dfrac{15}{19}$

Answer 2

Given Alice is selected there are $\binom{19}{4}$ to choose the remaining 4 people. Of these there are $\binom{18}{4}$ ways to make this choice without Bob. So I think the probability is

$\frac{\binom{18}{4}}{\binom{19}{4}} = 0.78947$

which is the same thing you got.