Given that x is a hypergeometric random variable, compute P(x) for each of the following cases.

a. N=5, n=3, r=3, x=1

b. N=9, n=5, r=3, x=3

c. N=4, n=2, r=2, x=2

d. N=4, n=2, r=2, x=0

The x is a hypergeometric random variable.

The Hypergeometric probability distribution is,

$$\begin{gathered} P\left(x\right)=\frac{\left(r \\ x\right)\left(N-r \\ n-x\right)}{\left(N \\ n\right)} \left[x=Maximum\left[0,n-\left(N-r\right)\right],...,Minimum\left(r,n\right)\right] \end{gathered}$$

The Hypergeometric probability for N=5, n=3, r=3, x=1 is computed as:

$$\begin{gathered} \begin{array}{rcl}P\left(x\right)& =& \frac{\left(3 \\ 1\right)\left(5-3 \\ 3-1\right)}{\left(5 \\ 3\right)}\\ & =& \frac{\frac{3!}{1!\left(3-1\right)!}\times \frac{2!}{2!\left(2-2\right)!}}{\frac{5!}{3!\left(5-3\right)!}}\\ & =& \frac{3\times 1}{10}\\ & =& 0.3\end{array} \end{gathered}$$

Hence, the required Hypergeometric probability for N=5, n=3, r=3, x=1 is 0.3.

The Hypergeometric probability for N=9, n=5, r=3, x=3 is computed as:

$$\begin{gathered} \begin{array}{rcl}P\left(x\right)& =& \frac{\left(3 \\ 3\right)\left(9-3 \\ 5-3\right)}{\left(9 \\ 5\right)}\\ & =& \frac{\frac{3!}{3!\left(3-3\right)!}\times \frac{6!}{2!\left(6-2\right)!}}{\frac{9!}{5!\left(9-5\right)!}}\\ & =& \frac{1\times 15}{126}\\ & =& 0.11905\end{array} \end{gathered}$$

Hence, the required Hypergeometric probability for N=9, n=5, r=3, x=3 is 0.11905.

The Hypergeometric probability for N=4, n=2, r=2, x=2 is computed as:

$$\begin{gathered} \begin{array}{rcl}P\left(x\right)& =& \frac{\left(2 \\ 2\right)\left(4-2 \\ 2-2\right)}{\left(4 \\ 2\right)}\\ & =& \frac{\frac{2!}{2!\left(2-2\right)!}\times \frac{2!}{0!\left(2-0\right)!}}{\frac{4!}{2!\left(4-2\right)!}}\\ & =& \frac{1\times 1}{6}\\ & =& 0.1666\end{array} \end{gathered}$$

Hence, the required Hypergeometric probability for N=4, n=2, r=2, x=2 is 0.1666.

The Hypergeometric probability for N=4, n=2, r=2, x=0 is computed as:

$$\begin{gathered} \begin{array}{rcl}P\left(x\right)& =& \frac{\left(2 \\ 0\right)\left(4-2 \\ 2-0\right)}{\left(4 \\ 2\right)}\\ & =& \frac{\frac{2!}{0!\left(2-0\right)!}\times \frac{2!}{2!\left(2-2\right)!}}{\frac{4!}{2!\left(4-2\right)!}}\\ & =& \frac{1\times 1}{6}\\ & =& 0.1666\end{array} \end{gathered}$$

Hence, the required Hypergeometric probability for N=4, n=2, r=2, x=0 is 0.1666.