Hypergeometric Probabilities

Posted on February 26, 2015 by Sumant Sumant

An box contains n balls, out of which r are red. We want to choose k balls, such that i of them are red.

Experiment: Picking k balls out of n balls where there are r balls of red color and rest different.

Sample space: \binom{n}{k}.

Event: Picking exactly i red balls.

We can choose i red balls out of r red balls in \binom{r}{i}. The remaining balls k-i have to be of different colors and have to be chosen from n-r and can be done in \binom{n-r}{k-i}. Therefore probability of choosing is \dfrac{\binom{r}{i} \binom{n-r}{k-i}}{\binom{n}{k}} for i \ge 0, satisfying i \le r,i \le k and k-i \le n-r. For all other i, the probability is zero.

The take home point is in Hypergeometric distribution unlike where the probability doesn’t change it changes here with each trial. Or one can think of urn problems without replacement and binomal as urn problems with replacement.
Note Hypergeomtric probability is also related to Vandermonde’s identity.
The Vandermonde’s identity is given by \binom{m+n}{k}=\sum\limits_{j=0}^{k}\binom{n}{j}\binom{m}{n-j}

About Sumant Sumant

I love Math and I am always looking forward to collaborate with fellow learners. If you need help learning math then please do contact me.
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