Hypergeometric Distribution

In binomial distribution, we examined the probability in the case of replacement or independency of the selections. However, without replacement we use hypergeometric distribution. Because of independency, for the binomial distribution the probability of selecting an item is constant for each try. For hypergeometric distribution the probability of selection changes after each succeding selection. The equation for hypergeometric distribution:

\begin{equation}\frac{C^s_x\;C^{N-s}_{n-x}}{C^N_n}\end{equation}

\begin{equation}N=Total\; Population\;\;\; s=Number \;of\;Success\;in\;Population\;\;\; \end{equation}

\begin{equation}x=Number\; of \;Sucess\;\;\; n=Number\;of\;Selections\;from\;Population\end{equation}

As you can understand, when we look at nominator we see the combination of selection of wanted object from population multiplied by the combination of selection of other object in population. In denominator, we can observe the possible combination of population objects.

For instance, 5 of 15 employees of a company will be selected for a company's important positions. If there 7 female and 8 male employees are there, what is the probability of selecting at least 3 men for this 5 position?

\begin{equation}p(x\ge3)=?\end{equation}

\begin{equation}N=15\;\;\;s=8\;\;\;x\ge3\end{equation}

\begin{equation}p(x\ge3)=p(x=3)+p(p=4)+p(x=5)\end{equation}

\begin{equation}p(x=3)=\frac{C^8_3\;C^7_2}{C^{15}_5}=0.3916084\end{equation}

\begin{equation}p(x=4)=\frac{C^8_4\;C^7_1}{C^{15}_5}=0.1631702\end{equation}

\begin{equation}p(x=5)=\frac{C^8_5\;C^7_0}{C^{15}_5}=0.01864802\end{equation}

\begin{equation}p(x\ge3)=0.3916084+0.1631702+0.01864802=0.5734266\end{equation}