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Mailrooms contaminated with anthrax. During autumn 2001, there was a highly publicized outbreak of anthrax cases among U.S. Postal Service workers. In Chance (Spring 2002), research statisticians discussed the problem of sampling mailrooms for the presence of anthrax spores. Let x equal the number of mailrooms contaminated with anthrax spores in a random sample of n mailrooms selected from a population of N mailrooms. The researchers showed that the probability distribution for x is given by the formula $\mathbf{P}\left(x\right)\mathbf{=}\frac{\left(\begin{array}{c}k\\ x\end{array}\right)\left(\begin{array}{c}N-k\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

where k is the number of contaminated mailrooms in the population. (In Section 4.4 we identify this probability distribution as the hypergeometric distribution.) Suppose N = 100, n = 3, and k = 20.

a. Find p(0).

b. Find p(1)

. c. Find p(2).

d. Find p(3)

Step by step solution

01

Given information

Here the distribution of xfollows hyper geometric distribution. The probability mass function ofx is $\mathrm{P}\left(\mathrm{x}\right)=\frac{\left(\begin{array}{c}\mathrm{k}\\ \mathrm{x}\end{array}\right)\left(\begin{array}{c}\mathrm{N}-\mathrm{k}\\ \mathrm{n}-\mathrm{x}\end{array}\right)}{\left(\begin{array}{c}\mathrm{N}\\ \mathrm{n}\end{array}\right)}$

02

Finding the value of  

a.

$$\begin{gathered} p\left(0\right)=\frac{\left(\begin{array}{c}20\\ 0\end{array}\right)\left(\begin{array}{c}100-20\\ 3-0\end{array}\right)}{\left(\begin{array}{c}100\\ 3\end{array}\right)} \\ =\frac{\left(\begin{array}{c}20\\ 0\end{array}\right)\left(\begin{array}{c}80\\ 3\end{array}\right)}{\left(\begin{array}{c}100\\ 3\end{array}\right)} \\ =0.508 \end{gathered}$$

Thus, the required value is 0.508.

03

Finding the value of p(1) 

b.

$$\begin{gathered} p\left(1\right)=\frac{\left(\begin{array}{c}20\\ 1\end{array}\right)\left(\begin{array}{c}100-20\\ 3-1\end{array}\right)}{\left(\begin{array}{c}100\\ 3\end{array}\right)} \\ =\frac{\left(\begin{array}{c}20\\ 1\end{array}\right)\left(\begin{array}{c}80\\ 2\end{array}\right)}{\left(\begin{array}{c}100\\ 3\end{array}\right)} \\ =0.3908 \end{gathered}$$

Thus, the required value is 0.3908.

04

Finding the value of p(2)

c.

$$\begin{gathered} p\left(1\right)=\frac{\left(\begin{array}{c}20\\ 2\end{array}\right)\left(\begin{array}{c}100-20\\ 3-2\end{array}\right)}{\left(\begin{array}{c}100\\ 3\end{array}\right)} \\ =\frac{\left(\begin{array}{c}20\\ 2\end{array}\right)\left(\begin{array}{c}80\\ 1\end{array}\right)}{\left(\begin{array}{c}100\\ 3\end{array}\right)} \\ =0.094 \end{gathered}$$

Thus, the required value is 0.94.

05

Finding the value of p(3)

d.

$$\begin{gathered} p\left(3\right)=\frac{\left(\begin{array}{c}20\\ 3\end{array}\right)\left(\begin{array}{c}100-20\\ 3-3\end{array}\right)}{\left(\begin{array}{c}100\\ 3\end{array}\right)} \\ =\frac{\left(\begin{array}{c}20\\ 3\end{array}\right)\left(\begin{array}{c}80\\ 0\end{array}\right)}{\left(\begin{array}{c}100\\ 3\end{array}\right)} \\ =0.0007 \end{gathered}$$

Thus, the required value is 0.0007.

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