Binomial Distribution Derivation

Feb 2010
163
15
in the 4th dimension
A population of \(\displaystyle n\) elements include \(\displaystyle np\) red ones and \(\displaystyle nq\) black ones, such that \(\displaystyle p+q=1\). A random sample of size \(\displaystyle r\) is taken with replacement. Prove that the probability of its including exactly \(\displaystyle k\) red elements is \(\displaystyle \binom{r}{k}p^{k}q^{r-k}\)
 

romsek

MHF Helper
Nov 2013
6,647
2,994
California
This derivation appears in a zillion places on the web but no.. you want one of us to painstakingly write it out for you.

This looks like it should answer your question.
 
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Walagaster

MHF Helper
Apr 2018
219
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Tempe, AZ
A population of \(\displaystyle n\) elements include \(\displaystyle np\) red ones and \(\displaystyle nq\) black ones, such that \(\displaystyle p+q=1\). A random sample of size \(\displaystyle r\) is taken with replacement. Prove that the probability of its including exactly \(\displaystyle k\) red elements is \(\displaystyle \binom{r}{k}p^{k}q^{r-k}\)
What a strange way to give the numbers of elements. So the probability of picking a red one in one trial is$$
P = \frac {\text{number of reds}}{\text{total number of elements}} = \frac{np}{np+nq}= \frac p {p+q} = \frac p 1 = p$$
Therefore you want the binomial B(n,p) distribution which, as romsek has pointed out, is all over the place on the net.
 
Feb 2010
163
15
in the 4th dimension
I might not have made my question clear, but I wanted to derive the distribution from a purely counting point of view.
So, what I wanted to ask actually was if there is a way to prove \(\displaystyle \frac{\binom{np}{k}\binom{nq}{r-k}}{n^r} = \binom{r}{k}p^{k}q^{r-k} \), which is where I got stuck.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
One way to get "r" red and "n- r" black is to get all r red first then n- r black:rrrrr...rrbbbb...bb. The probability of that is (p)(p)(p)…(p) (r times) times q(q)(q)…(q) (n- r times). The probability of that is p^rq^(n-r). But another way would be to get rrrr...rbrbbb..bb. The probability of that is (p)(p)(p)…(q)(p)(q)(q)…(q) which is exactly the same product. The number of different orders of r red and n- r black is the binomial coefficient \(\displaystyle \begin{pmatrix}n \\ r \end{pmatrix}\) so the probability of "r" red and "n-r" black in any order is \(\displaystyle \begin{pmatrix}n \\ r \end{pmatrix}p^rq^{n-r}\).