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If I have sit a lecture course with $17$ lectures and I study $10$ from $17$ and I when I sit the exam I answer $2$ questions from $4$ where only $1$ question can come from any given lecture what are the odds of me not having studied for two questions?

I've looked at card examples but become confused. I figured that the probability of getting one question to come up was $\frac{10}{17}\approx 0.58$. I also figured out that there are $6$ ways of choosing $2$ from $4$. I cant figure out how to combine the ideas to answer the question.

I thought it might be $\frac{10}{17}\times\frac{10}{17}\approx 0.3363$ and then multiply the answer by $2$.

I thought this because the odds of the events in the first two questions is the $0.58\times 0.58$ and the the odds of it occuring in the second two questions is $0.58\times 0.58$.

Some how this seems to leave something out?

1 Answers 1

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It is not entirely clear what is your question. I will try to give you a hint answering a question of what is the probability that, if you studied $10$ out of $17$ lectures, on an exam composed of $4$ questions from different lectures, you have not studied for $3$ or $4$.

First, suppose you have studied lectures $1$ through $10$ (this is without loss of generality since for any lectures you studied, I can renumber lectures such that you have studied $1$ to $10$).

Now suppose the test is composed of only one question. What is the probability you have studied for that question? It is $\frac{10}{17}$.

What if the test is composed of two questions. What is the probability you have studied for both? It is $\frac{10}{17}\cdot\frac{9}{16}\cdot\binom{2}{2}$. The first $\frac{10}{17}$ is the probability of choosing a question you have studied for out of the $17$. The second $\frac{9}{16}$ is the probability of choosing a question you have studied for, but now there are only $16$ questions to choose from, since questions from the same lecture cannot repeat, and out of those $16$ you have studied for $9$. And finally, there are $\binom{2}{2}$ ways to order two studied-for questions in two questions. Similarly, the probability you have studied for none is $\frac{7}{10}\cdot\frac{6}{16}\cdot\binom{2}{2}$. And the probability you have studied for one is $\frac{10}{17}\cdot\frac{7}{16}\cdot\binom{2}{1}$.

Final hint, when you get to your problem with a test composed of $4$ questions and you calculate the probability of not having studied for $3$ questions and the probability of not having studied for $4$ questions, the former is exactly $10$ times larger than the latter.