Chapter 14, Problem 14.41E

(a)

To determine

To find: The probability of exactly one infection among the 17 vaccinated children.

To obtain: The probability of exactly one infection among the 3 unvaccinated children.

To calculate: The probability of exactly one infection in each group.

Answer to Problem 14.41E

The probability of exactly one infection among the 17 vaccinated children is 0.3741.

The probability of exactly one infection among the 3 unvaccinated children is 0.0960.

The probability of exactly one infection in each group is 0.0359.

Explanation of Solution

Given info:

There is a group of 20 children in the nursery school who are exposed to whooping cough by playing with an infected child. Among these children, 17 have been vaccinated and 3 are unvaccinated.

Calculation:

The random variable “U” represents ‘the number of unvaccinated children who have been exposed to whooping cough will develop the infection’. Here, the random sample of 3(n) children has been randomly chosen as being exposed.

Also, there are two possible outcomes (develop the infections and do not develop the infections) with the probability of success ‘the chance that the unvaccinated children who been exposed to whooping will develop the infections (p) is 0.80’ and ‘the unvaccinated children who been exposed to whooping cough will not develop the infection is 0.20 $\left(=1-0.80\right)$’. Thus, U follows the binomial distribution with $n=3,p=0.80$

The random variable “V” represents ‘the number of vaccinated children who have been exposed to whooping cough will develop the infection’. Here, the random sample 17(n) children have been randomly chosen being exposed.

Also, there are two possible outcomes (develop the infections and do not develop the infections) with the probability of success ‘the chance that the vaccinated children who have been exposed to whooping cough will develop the infection (p) is 0.05’ and ‘the vaccinated children who been exposed to whooping cough will not develop the infection is 0.95 $\left(=1-0.05\right)$’. Thus, V follows the binomial distribution with $n=17,p=0.05$

The probability of exactly one infection among the 17 vaccinated children:

The probability using binomial distribution is given by,

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

Substitute n as 17, p as 0.05 and q as $0.95\left(=1-0.05\right)$

That is,

$\begin{array}{c}P\left(V=1\right)=\left(\begin{array}{l}17\\ 1\end{array}\right){\left(0.05\right)}^1{\left(0.95\right)}^{17-1}\\ =17\times 0.05\times 0.4401\\ =0.3741\end{array}$

Thus, the probability of exactly one infection among the 17 vaccinated children is 0.3741.

The probability of exactly one infection among the 3 unvaccinated children:

Substitute n as 3, p as 0.80 and q as $0.20\left(=1-0.80\right)$

That is,

$\begin{array}{c}P\left(U=1\right)=\left(\begin{array}{l}3\\ 1\end{array}\right){\left(0.80\right)}^1{\left(0.20\right)}^{3-1}\\ =3\times 0.80\times 0.04\\ =0.0960\end{array}$

Thus, the probability of exactly one infection among the 3 unvaccinated children is 0.0960.

The probability of exactly one infection in each group:

The events vaccinated children and unvaccinated children are independent of each other.

By the independent condition between the two events,

$\begin{array}{c}P\left(\begin{array}{l}1\text{ infection in }\\ \text{each group}\end{array}\right)=P\left(U=1\text{ and }V=1\right)\\ =P\left(U=1\right)\times P\left(V=1\right)\\ =0.0960\times 0.3741\\ =0.0359\end{array}$

Thus, the probability of exactly one infection in each group is 0.0359.

(b)

To determine

To find: The probability of exactly two infections in each group.

Answer to Problem 14.41E

The probability of exactly two infections in each group is 0.1978.

Explanation of Solution

Calculation:

The probability of two infections among the 17 vaccinated children:

Substitute n as 3, p as 0.80 and q as $0.95\left(=1-0.05\right)$ for the unvaccinated group.

Substitute n as 17, p as 0.05 and q as $0.95\left(=1-0.05\right)$ for the vaccinated group.

The possibilities are listed below:

$P\left(V=0\text{and}U=2\right),P\left(V=1\text{and}U=1\right),P\left(V=2\text{and}U=0\right)$

Consider,

$\begin{array}{c}P\left(V=0\text{and}U=2\right)=P\left(V=0\right)\times P\left(U=2\right)\\ =\left\{\left(\begin{array}{l}17\\ 0\end{array}\right){\left(0.05\right)}^0{\left(0.95\right)}^{17-0}\right\}\times \left\{\left(\begin{array}{l}3\\ 2\end{array}\right){\left(0.80\right)}^2{\left(0.20\right)}^{3-2}\right\}\\ =\left(1\times 1\times 0.4181\right)\times \left(3\times 0.128\right)\\ =0.4181\times 0.3840\end{array}$

$=0.1606$

Consider,

$\begin{array}{c}P\left(V=2\text{and}U=0\right)=P\left(V=2\right)\times P\left(U=0\right)\\ =\left\{\left(\begin{array}{l}17\\ 2\end{array}\right){\left(0.05\right)}^2{\left(0.95\right)}^{17-2}\right\}\times \left\{\left(\begin{array}{l}3\\ 0\end{array}\right){\left(0.80\right)}^0{\left(0.20\right)}^{3-0}\right\}\\ =\left(136\times 0.0025\times 0.4633\right)\times \left(1\times 1\times 0.008\right)\\ =0.1575\times 0.0080\end{array}$

$=0.0013$

Therefore,

$\begin{array}{c}P\left(\begin{array}{l}2\text{infections in }\\ \text{each group}\end{array}\right)=P\left(V=0\text{and}U=2\right)+P\left(V=1\text{and}U=1\right)+P\left(V=2\text{and}U=0\right)\\ =0.1606+0.0359+0.0013\\ =0.1978\end{array}$

Thus, probability of exactly two infections in each group is 0.1978.