Question

Given that xis a hypergeometric random variable, compute$\mathit{p}\left(x\right)$for each of the following cases:

a. N= 8, n= 5, r= 3, x= 2

b. N= 6, n= 2, r= 2, x= 2

c. N= 5, n= 4, r= 4, x= 3

Short answer

a. The probability distribution$p\left(x\right)$ is 0.5357.

b. The probability distribution$p\left(x\right)$ is 0.2.

c. The probability distribution$p\left(x\right)$ is 0.8.

Step-by-step solution

Given information

X is a hypergeometricrandom variable.

Computing the probability p(x) when N = 8, n = 5, r = 3, x = 2

For a hypergeometric random variable, the probability distribution

$p\left(x\right)=\frac{\left(\begin{array}{c}r\\ x\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

a.

Here$N=8,n=5,r=3,x=2$

$$\begin{gathered} p\left(x\right)=\frac{\left(\begin{array}{c}r\\ x\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)} \\ =\frac{\left(\begin{array}{c}3\\ 2\end{array}\right)\left(\begin{array}{c}8-3\\ 5-2\end{array}\right)}{\left(\begin{array}{c}8\\ 5\end{array}\right)} \\ =\frac{\left(\begin{array}{c}3\\ 2\end{array}\right)\left(\begin{array}{c}5\\ 3\end{array}\right)}{\left(\begin{array}{c}8\\ 5\end{array}\right)} \\ =\frac{3\times 10}{56} \\ =0.5357 \end{gathered}$$

Hence, the probability distribution$p\left(x\right)$ is 0.5357.

Computing the probability p(x) when N = 6, n = 2, r = 2, x = 2

b.

Here$N=5,n=4,r=4,x=3$

$$\begin{gathered} p\left(x\right)=\frac{\left(\begin{array}{c}r\\ x\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)} \\ =\frac{\left(\begin{array}{c}2\\ 2\end{array}\right)\left(\begin{array}{c}6-3\\ 2-2\end{array}\right)}{\left(\begin{array}{c}6\\ 2\end{array}\right)} \\ =\frac{\left(\begin{array}{c}3\\ 2\end{array}\right)\left(\begin{array}{c}4\\ 0\end{array}\right)}{\left(\begin{array}{c}6\\ 2\end{array}\right)} \\ =\frac{3\times 1}{15} \\ =0.2 \end{gathered}$$

Hence, the probability distribution$p\left(x\right)$ is 0.2.

Computing the probability p(x) when N = 5, n = 4, r = 3, x = 3

c.

Here,$N=5,n=4,r=4,x=3$

$$\begin{gathered} p\left(x\right)=\frac{\left(\begin{array}{c}r\\ x\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)} \\ =\frac{\left(\begin{array}{c}4\\ 3\end{array}\right)\left(\begin{array}{c}5-4\\ 4-3\end{array}\right)}{\left(\begin{array}{c}5\\ 4\end{array}\right)} \\ =\frac{\left(\begin{array}{c}4\\ 3\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)}{\left(\begin{array}{c}5\\ 4\end{array}\right)} \\ =\frac{4\times 1}{5} \\ =0.8 \end{gathered}$$

Hence, the probability distribution$p\left(x\right)$ is 0.8.