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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?

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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

$(1,2);(1;3);(1;4);(2,3);(2,4);(3,4)$

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

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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

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