Question

suppose $n = 11$ and $r = 4$. compute the hypergeometric probabilities for the following values of $n$ and $x$. if the calculations are not possible, please select
ot possible\ from below drop - downs, and enter \0\ in fields. round your answers, if necessary.
a. $n = 4, x = 1$ (to 2 decimals).

• select your answer -

b. $n = 2, x = 2$ (to 3 decimals).

• select your answer -

c. $n = 3, x = 0$ (to 4 decimals).

• select your answer -

d. $n = 5, x = 3$ (to 2 decimals).

• select your answer -

e. $n = 5, x = 5$ (to 2 decimals).

• select your answer -

Explanation:

Step1: Recall hypergeometric probability formula

The hypergeometric probability formula is $P(X = x)=\frac{\binom{r}{x}\binom{N - r}{n - x}}{\binom{N}{n}}$, where $N$ is the population size, $r$ is the number of successes in the population, $n$ is the sample size, and $x$ is the number of successes in the sample. Here $N = 11$, $r=4$.

Step2: Calculate part a

For $n = 4,x = 1$:
$\binom{r}{x}=\binom{4}{1}=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4$
$\binom{N - r}{n - x}=\binom{11 - 4}{4 - 1}=\binom{7}{3}=\frac{7!}{3!(7 - 3)!}=\frac{7\times6\times5}{3\times2\times1}=35$
$\binom{N}{n}=\binom{11}{4}=\frac{11!}{4!(11 - 4)!}=\frac{11\times10\times9\times8}{4\times3\times2\times1}=330$
$P(X = 1)=\frac{\binom{4}{1}\binom{7}{3}}{\binom{11}{4}}=\frac{4\times35}{330}\approx0.42$

Step3: Calculate part b

For $n = 2,x = 2$:
$\binom{r}{x}=\binom{4}{2}=\frac{4!}{2!(4 - 2)!}=\frac{4\times3}{2\times1}=6$
$\binom{N - r}{n - x}=\binom{11 - 4}{2 - 2}=\binom{7}{0}=1$
$\binom{N}{n}=\binom{11}{2}=\frac{11!}{2!(11 - 2)!}=\frac{11\times10}{2\times1}=55$
$P(X = 2)=\frac{\binom{4}{2}\binom{7}{0}}{\binom{11}{2}}=\frac{6\times1}{55}\approx0.109$

Step4: Calculate part c

For $n = 3,x = 0$:
$\binom{r}{x}=\binom{4}{0}=1$
$\binom{N - r}{n - x}=\binom{11 - 4}{3 - 0}=\binom{7}{3}=35$
$\binom{N}{n}=\binom{11}{3}=\frac{11!}{3!(11 - 3)!}=\frac{11\times10\times9}{3\times2\times1}=165$
$P(X = 0)=\frac{\binom{4}{0}\binom{7}{3}}{\binom{11}{3}}=\frac{1\times35}{165}\approx0.2121$

Step5: Calculate part d

For $n = 5,x = 3$:
$\binom{r}{x}=\binom{4}{3}=\frac{4!}{3!(4 - 3)!}=4$
$\binom{N - r}{n - x}=\binom{11 - 4}{5 - 3}=\binom{7}{2}=\frac{7!}{2!(7 - 2)!}=\frac{7\times6}{2\times1}=21$
$\binom{N}{n}=\binom{11}{5}=\frac{11!}{5!(11 - 5)!}=\frac{11\times10\times9\times8\times7}{5\times4\times3\times2\times1}=462$
$P(X = 3)=\frac{\binom{4}{3}\binom{7}{2}}{\binom{11}{5}}=\frac{4\times21}{462}\approx0.18$

Step6: Calculate part e

For $n = 5,x = 5$, since $r = 4$ (there are only 4 successes in the population), it is not possible to have 5 successes in the sample. So the probability is 0.

Answer:

a. 0.42
b. 0.109
c. 0.2121
d. 0.18
e. 0