Given that x is a Hypergeometric random variable with N=10, n=5, and r=6, compute the following:

a. $\mathit{P}\left(X=0\right)$

b. $\mathit{P}\left(x=1\right)$

c. $\mathit{P}\left(X⩽1\right)$

d. $\mathit{P}\left(X⩾2\right)$

The x is a Hypergeometric random variable with N=10, n=5, and r=6.

The Hypergeometric 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 is valid when $\left(n-x\right)⩽\left(N-r\right)$.

This means that the following condition is satisfied:

$0⩽x⩽r and 0⩽\left(n-x\right)⩽N-r$

i.e.,The Hypergeometric probability is valid when; $\max \left(0,n-N+r\right)⩽x⩽\min \left(r,n\right)$.

Here,$\left(n-x\right)>\left(N-r\right)\Rightarrow \left(5-0\right)>\left(10-6\right)$ .

Therefore,

The Hypergeometric probability $P\left(X=0\right)=0$with the values N=10, n=5, and r=6.

The probability$P\left(X=1\right)$ is calculated as:

$$\begin{gathered} \begin{array}{rcl}P\left(x=1\right)& =& \frac{\left(6 \\ 1\right)\left(10-6 \\ 5-1\right)}{\left(10 \\ 5\right)}\\ & =& \frac{\frac{6!}{1!\left(6-1\right)!}\times \frac{4!}{4!\left(4-4\right)!}}{\frac{10!}{5!\left(10-5\right)!}}\\ & =& \frac{6\times 1}{252}\\ & =& 0.02381\end{array} \end{gathered}$$

Hence, the required Hypergeometric probability $P\left(X=1\right)$for N=10, n=5, and r=6 is 0.02381.

The probability $P\left(x⩽1\right)$is calculated as:

$$\begin{gathered} P\left(X⩽1\right)=P\left(X=0\right)+P\left(X=1\right) \\ =\frac{\left(6l \\ 0\right)\left(10-6 \\ 5-0\right)}{\left(10 \\ 5\right)}+\frac{\left(6 \\ 1\right)\left(10-6 \\ 5-1\right)}{\left(10 \\ 5\right)} \\ =0+\frac{\frac{6!}{1!\left(6-1\right)!}\times \frac{4!}{4!\left(4-4\right)!}}{\frac{10!}{5!\left(10-5\right)!}} \\ =0+\frac{6\times 1}{252} \\ =0.02381 \end{gathered}$$

Where $P\left(X=0\right)=0$. Because $\left(n-x\right)⩽\left(N-r\right)\Rightarrow \left(5-0\right)>\left(10-6\right)$.

Hence, the required Hypergeometric probability$P\left(X⩽1\right)$ for N=10, n=5, r=6 is 0.02381.

The probability $P\left(X⩾2\right)$is computed as:

$$\begin{gathered} P\left(X⩾2\right)=1-P\left(X<2\right) \\ =1-\left(P\left(X=0\right)+P\left(X=1\right)\right) \\ =1-\left(\frac{\left(6 \\ 0\right)\left(10-6 \\ 5-0\right)}{\left(0 \\ 5\right)}+\frac{\left(6 \\ 1\right)\left(10-6 \\ 5-1\right)}{\left(10 \\ 5\right)}\right) \\ =1-\left(0+\frac{\frac{6!}{1!\left(6-1\right)!}\times \frac{4!}{4!\left(4-4\right)!}}{\frac{10!}{5!\left(10-5\right)!}}\right) \\ =1-\left(0+0.02381\right) \\ =\text{0.97619} \end{gathered}$$

Where$P\left(X=0\right)=0$. Because $\left(n-x\right)⩽\left(N-r\right)\Rightarrow \left(5-0\right)>\left(10-6\right)$.

Hence, the required Hypergeometric probability $P\left(X⩾2\right)$for N=10, n=5, r=6 is 0.97619.