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The exercise says: There are 35 women and 45 men in a school. From this population are selected first 10 students and after a second selection, a final group of 6 students is determined. What is the probability that in this last group there are at least 5 women?

At first I would say that it is hypergeometric, but it happens that it is like there is a sample of 6 within another sample of 10, and there I do not know what to do. Could you help me?

Thank you very much.

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The distribution is the same as if a group of 6 were selected directly from the full group.

The probability that a certain group (Alice, Bob, Carl, David, Edith, Francis) is chosen under the original scheme is $$\frac{\binom{74}{4}}{\binom{80}{10}} \cdot \frac{1}{\binom{10}{6}} = \frac{74!}{4! 70!} \cdot \frac{10! 70!}{80!} \cdot \frac{6!4!}{10!} = \frac{1}{\binom{80}{6}},$$ which is the same as the probability that this group is chosen if we had just selected a group of 6 directly.

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    That is to say that group of 10 is only to mislead? – 2017-02-27
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    Doesn't the OP want the probability of at least 5 women though? Seems to me like you just gave them the probability of choosing any 6 from a group. – 2017-02-27
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    @Heavenly96 What I did now was to define a random variable $X$ that counted the number of women who existed in a group of $6$ students out of a total of $80$ of whom $35$ are women. Then $X$ is hypergeometric. And finally calculate the probability that $X$ takes a value greater than or equal to $5$. Is it right for you what I did? – 2017-02-27
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    @emi That sounds correct to me. – 2017-02-27