I read that we can use hypergeometric distribution for finding the probability for without replacement cases because the probability of a particular event changes on every trial and binomial distribution fails.

I know how to solve questions using hypergeometric distribution but I don't understand how does using combinatorics for finding the probability helps in without replacement cases.

Can someone please tell me why hypergeometric distribution works for cases without replacement?


Let´s say you have $R$ (red) and $B$ (blue) elements, where $B+R=N$. Now you want to draw $r \in R $ elements and $b\in B$ without replacement. Again $r+b=n$

The number of ways to order $r$ drawn elements out of $R$ is $\binom{R}{r}$. For instance, if $R=4$ and $r=2$, then you have the following combinations to draw the two red elements out of $4$ red elements


Thus there are $6$ combinations which is equal to $\binom{R}{r}=\binom{4}{2}=6$

The same calculation can be done for the blue elements. To get the numbers to draw $r$ and $b$ elements (favorable cases, $\color{green}{F}$) the binomial coefficients has to be multiplied. We use that $B=N-R$ and $b=n-r$.

$$F=\binom{R}{r}\cdot \binom{N-R}{n-r}$$

Now we divide the term above by the number of all possible cases, $\color{violet}{P}$. Here you draw $r+b=n$ elements out of $N=R+B$ cases. You don´t care how much of the $n$ elements are blue or red, only $r\leq R$ and $b\leq B$ Thus the probability to draw $r$ and $b$ elements out of R and B elements is

$$\frac{\color{green}{F}}{\color{violet}{P}}=\frac{\binom{R}{r}\cdot \binom{N-R}{n-r}}{\binom{N}{n}}$$

  • $\begingroup$ Thanks for your answer. I understand how the formula is derived. I want to know how does hypergeometric correct the problem of binomial distribution for without replacement cases. $\endgroup$ – Rajesh R Jan 7 '18 at 17:36
  • 1
    $\begingroup$ You´re welcome. There is no direct link between the binomial distribution and the hypergeometric distribution. What you can say is that the probability to get a blue elements at the i-th drawing depends on the number of drawn blue and red elements at the drawings before. Thus the probability is not constant. $\endgroup$ – callculus Jan 7 '18 at 18:19
  • $\begingroup$ I took a very simple example to understand what's happening . For example, we have 2 red and 2 blue balls. By using traditional method, probability of picking 1 red and 1 blue balls without replacement is $\frac {2*2}{4*3} = \frac{4}{12}$ but with hypergeometric, the answer is $\frac {2*2}{4C2} = \frac {4}{6}$. Why are the answers different? $\endgroup$ – Rajesh R Jan 7 '18 at 18:48
  • 1
    $\begingroup$ @RajeshR You have two ways to draw $1\color{red}r$ and $1\color{blue}b$: $\color{red}r\color{blue}b$ and $\color{blue}b\color{red}r$-both with the same probability. The probability to draw $rb$ is $\frac{\color{red}r}{\color{red}r+\color{blue}b}\cdot \frac{\color{blue}b}{\color{red}r+\color{blue}b-1}=\frac24\cdot \frac{2}{3}=\frac{4}{12}$. Since you have two ways the probability is $2\cdot \frac{4}{12}=\frac46=\frac23$. This is the same result you have calculated with the $\texttt{hypergeometric distribution}$. $\endgroup$ – callculus Jan 7 '18 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.