Pinworm Infestation. Use Procedure 5.1 on page 236 to solve part (g) of Exercise 5.165.

The given statement from Exercise 5.165 is:

Pinworm infestation, which is commonly found in children, can be treated with the drug pyrantel pamoate.

According to the Merck Manual, the treatment is effective in 90% of cases. Suppose that three children with pinworm infestation are given pyrantel pamoate.

First, identify the success:

Treatment with pyrantel pamoate cures a person with pinworm.

Now determine the probability of success (p):

In 90% of situations, the likelihood of a person being healed by treatment is high.

Therefore, $p=0.9$.

Determine the number of trials (n).

Given that pyrantel pamoate is given to three infants with pinworm infestation. Therefore,$n=3$.

The formula is:

$P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)P^x{\left(1-p\right)}^{n-x}$

Insert the values in the formula,

$P\left(X=x\right)=\left(\begin{array}{c}3\\ x\end{array}\right){\left(0.9\right)}^x{\left(1-0.9\right)}^{3-x}$

We can see that X is a binomial random variable with parameters and the binomial distribution $p=0.9\mathrm{and}n=3$.

We must now calculate the probabilities of all Xvalues. That is, Xis a random variable with values ranging from 0 to 3.

$$\begin{gathered} P\left(X=0\right)=\left(\begin{array}{c}3\\ 0\end{array}\right){\left(0.9\right)}^0{\left(0.1\right)}^3 \\ ={\left(0.1\right)}^3 \\ =0.001 \\ P\left(X=1\right)=\left(\begin{array}{c}3\\ 1\end{array}\right){\left(0.9\right)}^1{\left(0.1\right)}^2 \\ =3\times 0.9\times {\left(0.1\right)}^2 \\ =0.0270 \\ P\left(X=2\right)=\left(\begin{array}{c}3\\ 2\end{array}\right){\left(0.9\right)}^2{\left(0.1\right)}^1 \\ =3\times {\left(0.9\right)}^2\times 0.1 \\ =0.243 \\ P\left(X=3\right)=\left(\begin{array}{c}3\\ 3\end{array}\right){\left(0.9\right)}^0{\left(0.1\right)}^3 \\ =1\times {\left(0.9\right)}^3 \\ =0.7290 \end{gathered}$$