Step 1

Given:

Let,

\(\displaystyle{X}=\text{Number of American adult have had chickenpox by the time they reach adulthood}.\)

\(\displaystyle{p}=\text{probability of American adult have had chickenpox by the time they reach adulthood, we have}\)

\(\displaystyle{p}={0.90}\)

\(\displaystyle{X}\sim{B}in{o}{m}{i}{a}{l}{\left({n},{p}\right)}\)

\(\displaystyle{Y}=\text{Number of American adult have not had chickenpox by the time they reach adulthood}.\)

\(\displaystyle{q}=\text{probability of American adult have not had chickenpox by the time they reach adulthood, we have}\)

\(\displaystyle{q}={1}-{p}={0.10}\)

\(\displaystyle{Y}\sim{B}in{o}{m}{i}{a}{l}{\left({n},{q}\right)}\)

Step 2

Solution:

(d) Let, \(\displaystyle{n}={10}\)

\(\displaystyle{X}\sim{B}in{o}{m}{i}{a}{l}{\left({n}={10},{p}={0.9}\right)}\)

Now,

\(\displaystyle{P}{\left({X}\geq{1}\right)}={1}-{P}{\left({x}={0}\right)}\)

\[=1-(\begin{array}{c}10\\ 0\end{array})(0.9)^{0}(0.1)^{10}\]

\(\displaystyle={0.9999}\approx{1}\)

\(\displaystyle{P}{\left({X}\geq{1}\right)}={1}\)

(e) Let,

\(\displaystyle{Y}\sim{B}in{o}{m}{i}{a}{l}{\left({n}={10},{q}={0.1}\right)}\)

Now,

\(\displaystyle{P}{\left({X}\le{3}\right)}={P}{\left({x}={0}\right)}+{P}{\left({x}={1}\right)}+{P}{\left({x}={2}\right)}={P}{\left({x}={3}\right)}\)

\[=(\begin{array}{c}10\\ 0\end{array})(0.1)^{0}(0.9)^{10}+(\begin{array}{c}10\\ 1\end{array})(0.1)^{1}(0.9)^{9}+(\begin{array}{c}10\\ 2\end{array})(0.1)^{2}(0.9)^{8}+(\begin{array}{c}10\\ 3\end{array})(0.1)^{3}(0.9)^{7}\]

\(\displaystyle{P}{\left({X}\le{3}\right)}={0.9872}\)