If you know 2 of 8 questions but the test has only 4 questions. What is the probability you will know at least one of the questions on the test? Background:
This is a hypothetical question that and my friend and I have been arguing over all day. We're both upper-year mathematics students and should reasonably be able to come up with the solution, however, we've been arguing about it non stop.
Situation:
There's a test bank with $8$ questions in it. The test will randomly give you $4$ of these questions. If you happen to know $2$ of the $8$ questions ahead of time,
A) What is the probability your test would have at least one of the questions you know?
B) What is the probability your test would have both of the questions you know?
So far we've come up with the solutions for part A) of:

*

*simply $25\%$ because you know $25$ percent of the questions

*$39 \%$ because you have a $\frac{2}{8}$ chance of knowing the first question and a $\frac{1}{7}$ of the second and $\frac{2}{8}+ \frac{1}{7} \approx 0.39$

*$\frac{1}{16}$ because the test has $\frac{1}{2}$ of the total questions and you have a $\frac{1}{8}$ chance of getting your question

*$\frac{1}{8}$ similarly, because you know two of the questions so $2 \times \frac{1}{16}= \frac{1}{8}$
For part B), we think it could be

*

*$\frac{2}{8} \times \frac{1}{7} \approx 3\% $

*$\frac{1}{16}$?

We're really lost and at this point, I think we need a third party to arbitrate our discussion haha.

I'd like to offer an alternative approach. I will instead find the probability that neither of the two questions you know is selected, then subtract that from 1. There are $6\choose 4$ ways to pick a set of 4 questions from the 6 questions you don't know, and there are $8\choose 4$ ways to pick a set of 4 questions without restrictions. So, the probability that you know none of the 4 questions is
$$\frac{6\choose 4}{8\choose 4}=\frac{3}{14}\approx 21.43\%$$
Subtracting this from 1 gives a probability of $\frac{11}{14}$ of knowing at least 1 question on the test, or about a $78.57\%$ chance.
I checked this with python, and it numerically lines up.
