# Probability that at least 3 of the 5 randomly selected questions can be solved

As my friend my studying for her final for her Philosophy class, she asked me a question.

The professor assigned a list of 10 essay questions, and would be picking 5 to put on the exam. Of the 5 that he put on the exam, you only needed to answer 3. If she only knew how to answer 4 of the essay questions, what is the probability that she would be able to be solve at least 3 out of the 5 questions given to her? How does this probability change if she knew how to answer 5 of them? 6?

Thanks so much for the help!

• edit the post to include what have you tried? also use the self-study tag. – Siong Thye Goh Dec 15 '18 at 11:48
• Use hypergeometric distribution: An urn contains 4 green balls and 6 red balls. Five balls are selected at random without replacement. What is the probability that at least 3 green balls are selected? In R, sum(dhyper(3:4, 4,6, 5)) returns 0.2619048. Your friend needs to work a little harder. – BruceET Dec 15 '18 at 21:32

! (1) From this description there are $$N=10$$ practice questions and I answered $$K=4$$ of them correctly in my practice attempts. There are $$n=5$$ random questions assigned to the exam, so the probability I answer at least three of the exam questions correctly is:\begin{aligned} \mathbb{P}(\text{Answer at least } 3 \text{ correctly}) &= \sum_{k=3}^5 \text{Hypergeom}(k|N=10 ,K=4, n=5) \\[6pt] &= \sum_{k=3}^5 \frac{{4 \choose k} {6 \choose 5-k}}{{10 \choose 5}} \\[6pt] &= \frac{{4 \choose 3} {6 \choose 2}}{{10 \choose 5}} + \frac{{4 \choose 4} {6 \choose 1}}{{10 \choose 5}} + 0 \\[6pt] &= \frac{4 \cdot 15}{252} + \frac{1 \cdot 6}{252} \\[6pt] &= \frac{66}{252} = 0.2619048. \\[6pt] \end{aligned} (2) In answering Part (1) we treated all the assigned exam questions as random. However, suppose instead that having read this question, I know that it was one of the ones I got wrong in the practice attempt. (Note that the question assumes I have not read the other questions on the exam, but I have to read this one in order to answer it!) In that case there would be $$N=9$$ remaining practice questions and I answered $$K=4$$ of them correctly in my practice attempts. There are $$n=4$$ random questions assigned to the exam (setting aside this one), so the probability I answer at least three of the exam questions correctly is:\begin{aligned} \mathbb{P}(\text{Answer at least } 3 \text{ correctly}) &= \sum_{k=3}^5 \text{Hypergeom}(k|N=9 ,K=4, n=4) \\[6pt] &= \sum_{k=3}^5 \frac{{4 \choose k} {5 \choose 4-k}}{{9 \choose 4}} \\[6pt] &= \frac{{4 \choose 3} {5 \choose 1}}{{9 \choose 4}} + \frac{{4 \choose 4} {5 \choose 0}}{{9 \choose 4}} + 0 \\[6pt] &= \frac{4 \cdot 5}{126} + \frac{1 \cdot 1}{126} \\[6pt] &= \frac{21}{126} = 0.1666667. \\[6pt] \end{aligned} Now, looking at this answer a logical paradox arises. If the answer is correct then the assumption that it is answered incorrectly is wrong, and so we should not have conditioned on this, and so the answer is actually incorrect. On the other hand, if this answer is incorrect (as assumed) then the above calculations are correct, so the answer is actually correct. Thus, we obtain a situation in which the answer is correct if it is incorrect, and incorrect if it is correct. This is somewhat similar to Russell's paradox.