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Suppose $X$ has a hypergeometric distribution with
$N=20, n=4,$ and $K=4 .$ Determine the following:

$\begin{array}{ll}{\text { (a) } P(X=1)} & {\text { (b) } P(X=4)} \\ {\text { (c) } P(X \leq 2)} & {\text { (d) Determine the mean and variance of } X .}\end{array}$

$x \rightarrow$ hypergeometric random variable

$N=20 \quad n=4 \quad k=4$

(a) $P(x=1)$

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

$P(x=1)=\frac{\left(\begin{array}{l}{4} \\ {1}\end{array}\right)\left(\begin{array}{l}{16} \\ {3}\end{array}\right)}{\left(\begin{array}{c}{20} \\ {4}\end{array}\right)}=0.4623$

(b) $P(x=4)=\frac{\left(\begin{array}{l}{4} \\ {4}\end{array}\right)\left(\begin{array}{l}{16} \\ {0}\end{array}\right)}{\left(\begin{array}{l}{20} \\ {4}\end{array}\right)}=0.00021$

(c) $P(x \leq 2)=P(x=0)+P(x=1)+P(x=2)$

$=\frac{\left(\begin{array}{l}{4} \\ {0}\end{array}\right)\left(\begin{array}{l}{16} \\ {4}\end{array}\right)}{\left(\begin{array}{l}{20} \\ {4}\end{array}\right)}+\frac{\left(\begin{array}{l}{4} \\ {1}\end{array}\right)\left(\begin{array}{l}{16} \\ {3}\end{array}\right)}{\left(\begin{array}{l}{20} \\ {4}\end{array}\right)}+\frac{\left(\begin{array}{l}{4} \\ {2}\end{array}\right)\left(\begin{array}{l}{16} \\ {2}\end{array}\right)}{\left(\begin{array}{l}{20} \\ {4}\end{array}\right)}=0.9866$

(d) $\mu=E(x)=n P$

$=4 \frac{k}{N}=4 \times \frac{4}{20}=0.8$

$\sigma^{2}=n P(1- P)\left(\frac{N-n}{N-1}\right)$

$=4 \times 0.2 \times 0.8 \times \frac{16}{19}=0.539$

A company employs 800 men under the age of $55 .$
Suppose that $30\%$ carry a marker on the male chromosome
that indicates an increased risk for high blood pressure.

(a) If 10 men in the company are tested for the marker in this
chromosome, what is the probability that exactly 1 man
has the marker?
(b) If 10 men in the company are tested for the marker in this
chromosome, what is the probability that more than 1 has
the marker?

$N=800$

$P=30\% \rightarrow$ high blood pressure

(a) $n=10 \quad N=800$

$k=0.3 \times 800=240$

$P(x=1)=\frac{\left(\begin{array}{c}{240} \\ {1}\end{array}\right)\left(\begin{array}{c}{800-240} \\ {10-1}\end{array}\right)}{\left(\begin{array}{c}{800} \\ {10}\end{array}\right)}=0.1201$

(b) $P(x>1) \quad n=10$

$P(x>1)=1-P(x \leq 1)$

$P(x>1)=1-[P(x=1)+P(x=0)]$

$=1-0.1201-\frac{\left(\begin{array}{c}{240} \\ {0}\end{array}\right)\left(\begin{array}{c}{800-240} \\ {10}\end{array}\right)}{\left(\begin{array}{c}{800} \\ {10}\end{array}\right)}$

$=0.8523$