Question

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?

Answer 1

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

Answer 2

\[P(1sf)=\frac{\left(\begin{matrix}5 \\ 1\end{matrix}\right)\left(\begin{matrix}6 \\ 3\end{matrix}\right)}{\left(\begin{matrix}11 \\ 4\end{matrix}\right)}\]
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.
\[P(3mystery+1sf)=(\frac{100\times 24}{11\times 10\times 9\times 8})^{2}\]