Question

Whooping cough (pertussis) is a highly contagious bacterial infection that was a major cause of childhood deaths before the development of vaccines. About 80 % of unvaccinated children who are exposed to whooping cough will develop the infection, as opposed to only about 5% of vaccinated children. A group of 20 children at a nursery school is exposed to whooping cough by playing with an infected child.
(a) If all 20 have been vaccinated, what is the mean number of new infections?
(b) What is the probability that no more than 2 of the 20 children develop infections if all 20 have been vaccinated?

Answer 1

Answer:

(a) the mean number for new infections is equal to 1

(b) the probability that no more than 2 of the 20 children develop infections is 0.9245

Step-by-step explanation:

If all 20 have been vaccinated, the number of new infection follows a binomial distribution, so, the probability that x children develop infections is calculated as:

$p(x)=nCx*p^{x}*(1-p)^{n-x}$

$nCx=\frac{n!}{x!(n-x)!}$

where n is the number of children and p is the probability to develop the infection given that all 20 have been vaccinated. Additionally, the mean number E(x) of new infections is equal to:

$E(x)=n*p$

Now, taking into account that all children are vaccinated, we can replace n by 20 and p by 0.05, so the mean number of new infections is:

$E(x)=20*0.05=1$

Then, the probability that no more than 2 of the 20 children develop infections is calculated as:

$p(x\leq 2)=p(0)+p(1)+p(2)$

where p(x) is:

$p(x)=20Cx*0.05^{x}*(1-0.05)^{20-x}$

So, p(0), p(1) and p(2) are equal to:

$\\p(0)=20C0*0.05^{0}*(1-0.05)^{20-0}=0.3585\\p(1)=20C1*0.05^{1}*(1-0.05)^{20-1}=0.3773\\p(2)=20C2*0.05^{2}*(1-0.05)^{20-2}=0.1887$

Finally, $p(x\leq 2)$ is equal to:

$p(x\leq 2)=0.3585+0.3773+0.1887=0.9245$