Question

A random variable, \(x\), has a hypergeometric distribution with \(\mathrm{N}=13, X=9\), and \(n=4\).

a. Calculate \(\mathrm{P}(\mathrm{x}=3)\).

b. Calculate \(\mathrm{P}(\mathrm{x}=6)\).

c. Calculate \(\mathrm{P}(x \geq 4)\).

d. Find the largest \(x^{\prime}\) so that \(P\left(x>x^{\prime}\right)>0.25\).

a. The probability is (Round to four decimal places as needed.)

Answer 1

Solution: We know the pmf of geometric distribution

$p(x)=\frac{\binom{X}{x}\binom{N-X}{n-x}}{\binom{N}{n}}$

Where $x\leq n;x\leq X;n-x\leq N-X$

Now,

(1) P(x=3)

$=\frac{\binom{9}{3}\binom{13-9}{4-3}}{\binom{13}{4}}$

$=0.4699$

(b) P(X=6)=0 not satisfy the conditions

(c)

$P(x\geq 4)=1-P(x<4)$

                     $=1-\left [ P(x=0)+P(x=1)+P(2)+P(x=3) \right ]$

                      $=1-\left [ \frac{\binom{9}{0}\binom{13-9}{4-0}}{\binom{13}{4}}+\frac{\binom{9}{1}\binom{13-9}{4-1}}{\binom{13}{4}}+\frac{\binom{9}{2}\binom{13-9}{4-2}}{\binom{13}{4}}+\frac{\binom{9}{3}\binom{13-9}{4-3}}{\binom{13}{4}} \right ]$

                       $=0.17622$

(d) Largest of x' is 4

since

higest possible = min{X, n}=min{9,4}=4