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There are $100$ questions, and $10$ of them are chosen randomly for an exam. For each choice $10$ questions there is the same probability. A student has managed to get $60$ of the $100$ questions and thus know only to solve them.

To pass the exam he needs to solve correctly at least $6$ of the $10$ questions.

What is the probability that the student will pass the exam?

I thinkt it's ${{6+7+8+9+10} \over 60} = {2 \over 3}$. Is that correct?

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This problem is about the hypergeometric distribution. This distribution is used when you have a population of $N$ objects containing $K$ "successes", and you draw $n$ of the objects without replacement. Then the number of successes in your draw is a random variable $X$. It turns out that

$$P(X=k)=\frac{{K \choose k} {N-K \choose n-k}}{ {N \choose n}}$$

where the parentheses denote a binomial coefficient. The numerator is the number of ways to choose $k$ successes out of the total of $K$ successes times the number of ways to choose $n-k$ failures out of the total of $N-K$ failures. The denominator is the total number of ways to choose $n$ things from the whole population.

You then want $P(X \geq 6)$ which is $P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)$.

As an aside, when $k$ is much less than $K$ and $n-k$ is much less than $N-K$, the distribution of $X$ can be approximated by Binomial(n,K/N), that is, $P(X=k) \approx {n \choose k} (K/N)^k (1-K/N)^{n-k}$. This is exactly what you would get if the draws were done with replacement. In this case this approximation is pretty good.

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