## Related questions with answers

Each year a college awards five merit-based scholarships to members of the entering freshman class who have exceptional high school records. The initial pool of applicants for the upcoming academic year has been reduced to a “short list” of eight men and ten women, all of whom seem equally deserving. If the awards are made at random from among the eighteen finalists, what are the chances that both men and women will be represented?

Solution

Verified$\textit{The probability that all five awards are given to women:}$

Let $X$ be the number of women who got an award.

$X$ is a hypergeometric variable with parameters $N=18,\ r=10,\ w=8,\ n=5$, assuming the meanings from Therem 3.2.2.

The probability that $k=5$ women get an award is given by the same theorem:

$\begin{equation*} P(X=5)=\dfrac{\binom{10}{5}\binom{8}{0}}{\binom{18}{5}}=\dfrac{\dfrac{10!}{5!\cdot 5!}\cdot \dfrac{8!}{0!\cdot 8!}}{\dfrac{18!}{5!\cdot 13!}}=0.0294 \end{equation*}$

[1ex]

$\textit{The probability that all five awards are given to men:}$

Same $X$ as above will be used, only $k$ changes.

The probability that $k=0$ women get an award is given by:

$\begin{equation*} P(X=0)=\dfrac{\binom{10}{0}\binom{8}{5}}{\binom{18}{5}}=\dfrac{\dfrac{10!}{0!\cdot 10!}\cdot \dfrac{8!}{5!\cdot 3!}}{\dfrac{18!}{5!\cdot 13!}}=\dfrac{56}{8568}=0.0065 \end{equation*}$

[1ex]

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