# What Is a Hypergeometric Probability Distribution?

Rule

### CriteriaforUsageoftheHypergeometricDistribution

A random variable $X$ is hypergeometrically distributed if it meets the following criteria:

• You draw a random element from a population that is divided into two or more groups: Group $A$, group $B$, group $C$ and so on.

• You draw $r$ elements, without replacing any of them, distributed between these groups.

• $X$ is the number of elements you draw from the different groups.

Theory

### TheHypergeometricDistribution

The probability of drawing $k$ elements from group $A$ and $r-k$ elements from group $B$ is given by this formula:

where the different letters in the formula are:

• $k$ is the number of elements you draw from group $A$.

• $r$ is the total number of elements you draw from both group $A$ and $B$.

• This makes $r-k$ the number of elements you draw from group $B$.

• $a$ is the number of elements in $A$, and $b$ is the number of elements in $B$.

• $n=a+b$ is the total number of elements in the set.

Example 1

You draw five cards from a normal deck of cards and want to see how many hearts you get. Let $X$ be the number of hearts among your cards. What is the probability of getting three hearts?

This is the information you can deduce from the text: The population is 52, as there are 52 cards in a deck of cards. The relevant groups in this case are hearts and non-hearts. There are 13 hearts and $52-13=39$ non-hearts in the deck. You’ve drawn a total of five cards, with three of them being hearts. That means $5-3=2$ of your cards are non-hearts.

After sorting through this information, you can see that $X$ is hypergeometrically distributed, and you can insert the numbers you found into the above formula. That gives you

The probability of drawing three hearts is $8.2$ %.

Example 2

A box of light bulbs contains 100 light bulbs. Seven of them are broken. If you draw four random bulbs from the box, what is the probability of drawing only functioning light bulbs?

You can see that the population is 100 from the text. There are two groups, one group containing the broken light bulbs and another containing the functioning ones. The group of broken light bulbs has seven elements, while the group of functioning light bulbs has $100-7=93$ elements. You want to draw $k=4$ light bulbs. From this you understand that $X$ is hypergeometrically distributed, and that you want to find the probability of $X=0$. You use the formula from before and get that

 $P\phantom{\rule{-0.17em}{0ex}}\left(X=0\right)=\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{7}{0}\right)\cdot \left(\genfrac{}{}{0.0pt}{}{93}{4}\right)\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{100}{4}\right)\phantom{\rule{0.17em}{0ex}}}\approx 0.745.$

This means there’s a $74.5$ % chance that all four of the drawn light bulbs work.

Example 3

Danny Boyle is setting up a new musical in the West End and is holding auditions for extras. There are 14 female and 7 male actors at this audition, and Boyle needs 8 actors regardless of gender. Imagine every actor auditioned equally well, so the probability of getting the part is the same for everyone.

What’s the probability of four actors of each gender getting the parts?

What’s the probability of all the male actors getting a part?

What’s the probability that at least two of the chosen actors are female?

You can see that group $A$ (the women) has $14$ elements and that group $B$ (the men) has $7$ elements. This gives you $14+7=21$ elements in total. Danny needs to pick $r=8$ of them. Let $X$ be the number of women he picks.

1.
Here you’re interested in the probability of drawing 4 from group $A$ and 4 from group $B$. That means you have to put $k=4$ and $r=8$ into the formula, giving you
$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(X=4\right)& =\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{14}{4}\right)\left(\genfrac{}{}{0.0pt}{}{7}{8-4}\right)\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{21}{8}\right)\phantom{\rule{0.17em}{0ex}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 0.172.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(X=4\right)=\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{14}{4}\right)\left(\genfrac{}{}{0.0pt}{}{7}{8-4}\right)\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{21}{8}\right)\phantom{\rule{0.17em}{0ex}}}\approx 0.172.& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

The probability of Danny picking equal amounts women and men is around $17.2$ %.
2.
In this task, you’re interested in the probability of drawing all the elements in group $B$. As group $B$ has 7 elements, and Danny needs 8 extras, he has to pick exactly 1 woman as well. This makes $P\phantom{\rule{-0.17em}{0ex}}\left(X=1\right)$ the probability you’re looking for. You use the same formula as in the previous part, giving you
$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(X=1\right)& =\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{14}{1}\right)\left(\genfrac{}{}{0.0pt}{}{7}{7}\right)\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{21}{8}\right)\phantom{\rule{0.17em}{0ex}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =6.9\cdot 1{0}^{-5},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(X=1\right)=\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{14}{1}\right)\left(\genfrac{}{}{0.0pt}{}{7}{7}\right)\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\left(\genfrac{}{}{0.0pt}{}{21}{8}\right)\phantom{\rule{0.17em}{0ex}}}=6.9\cdot 1{0}^{-5},& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

which means the probability of Danny picking seven men and one woman is around $0.007$ %.
3.
Here you want to calculate the probability of ending up with at least two women. This is the compliment of the event “Danny picks 0 women or 1 woman”, the probability of which is $P\phantom{\rule{-0.17em}{0ex}}\left(X=0\right)+P\phantom{\rule{-0.17em}{0ex}}\left(X=1\right)$.

$P\phantom{\rule{-0.17em}{0ex}}\left(X=0\right)$ is 0, because there being only seven men makes it so that Danny always have to pick at least one woman.

You found $P\phantom{\rule{-0.17em}{0ex}}\left(X=1\right)$ in the previous part. That means that

This makes the probability of Boyle picking two or more women $99.9$ %.

Example 4

Bradley Cooper goes to a buffet to have some sushi. There is space for 15 pieces on his plate, and he’s determined to fill it up. In the buffet there is 30 pieces of salmon maki, 20 pieces of tempura maki and 15 pieces of maki with scallops. What’s the probability of him choosing five pieces of each?

Bradley needs to choose between three different groups. He needs to pick 5 pieces from the first group of 30, from the second group of 20 pieces he also wants 5, and the same is the case with the final group of 15 pieces. All together, that is $30+20+15=65$ pieces in the buffet. Then you can insert the numbers straight into the formula for the hypergeometric distribution and get this:

The probability of Bradley picking five pieces of each type is $3.2$ %.