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OpenStudy (anonymous):

Ellyssa has 6 mystery books and 5 science fiction books on her shelf. She takes down 4 books at random. What is the probability that she has 3 mysteries and 1 science fiction book?

OpenStudy (kropot72):

The hypergeometric distribution applies (sampling without replacement) \[P(3mystery)=\frac{\left(\begin{matrix}6 \\ 3\end{matrix}\right)\left(\begin{matrix}5 \\ 1\end{matrix}\right)}{\left(\begin{matrix}11 \\ 4\end{matrix}\right)}\] \[P(3mystery)=\frac{100\times 24}{11\times 10\times 9\times 8}\]

OpenStudy (kropot72):

\[P(1sf)=\frac{\left(\begin{matrix}5 \\ 1\end{matrix}\right)\left(\begin{matrix}6 \\ 3\end{matrix}\right)}{\left(\begin{matrix}11 \\ 4\end{matrix}\right)}\] The probability of 1 SiFi book in a random selection of 4 books is the same as the probability of 3 mysteries in a random selection of 4 books. Therefore the probability of a random selection of 4 books having 3 mysteries and 1 SiFi is the product of the probabilities of the two sub-events. \[P(3mystery+1sf)=(\frac{100\times 24}{11\times 10\times 9\times 8})^{2}\]

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