Let X be the number of men will be alive and n be sample number of men.

From the given information, probability that a men will be alive in 30 years is \(\displaystyle{\frac{{{2}}}{{{3}}}}\ {\quad\text{and}\quad}\ {n}={5}\).

Here, men are independent and probability of success is constant. Hence, X follows binomial distribution with parameters \(\displaystyle{n}={5}\ {\quad\text{and}\quad}\ {p}={\frac{{{2}}}{{{3}}}}\).

The probability mass function of binomial random variable X is

\[P(X=x)=(\begin{array}{c}n\\ x\end{array})p^{x}(1-p)^{n-x}; x=0,1,...,n\]

Step 3

a. The probability that in 30 years all 5 men will be alive is

\[P(X=5)=(\begin{array}{c}5\\ 5\end{array})(\frac{2}{3})^{5} (1-\frac{2}{3})^{5-5}\]

\(\displaystyle={0.1317}\)

Thus, the probability that in 30 years all 5 men will be alive is 0.1317.

Step 4

b. The probability that in 30 years at least 3 men will be alive is

\(\displaystyle{P}{\left({X}\geq{3}\right)}={P}{\left({X}={3}\right)}+{P}{\left({X}={4}\right)}+{P}{\left({X}={5}\right)}\)

\[=(\begin{array}{c}5\\ 3\end{array})(\frac{2}{3})^{3} (1-\frac{2}{3})^{5-3}+(\begin{array}{c}5\\ 4\end{array})(\frac{2}{3})^{4} (1-\frac{1}{2})^{5-4}+(\begin{array}{c}5\\ 5\end{array})(\frac{2}{3})^{5} (1-\frac{2}{3})^{5-5}\]

\(\displaystyle={0.3292}+{0.3292}+{0.1317}\)

\(\displaystyle={0.7901}\)

The probability that in 30 years at least 3 men will be alive is 0.7901.

Step 5

c. The probability that in 30 years all only 2 men will be alive is

\[P(X=2)=(\begin{array}{c}5\\ 2\end{array})(\frac{2}{3})^{2} (1-\frac{2}{3})^{5-2}\]

\(\displaystyle={0.1646}\)

Thus, the probability that in 30 years all only 2 men will be alive is 0.1646.

From the given information, probability that a men will be alive in 30 years is \(\displaystyle{\frac{{{2}}}{{{3}}}}\ {\quad\text{and}\quad}\ {n}={5}\).

Here, men are independent and probability of success is constant. Hence, X follows binomial distribution with parameters \(\displaystyle{n}={5}\ {\quad\text{and}\quad}\ {p}={\frac{{{2}}}{{{3}}}}\).

The probability mass function of binomial random variable X is

\[P(X=x)=(\begin{array}{c}n\\ x\end{array})p^{x}(1-p)^{n-x}; x=0,1,...,n\]

Step 3

a. The probability that in 30 years all 5 men will be alive is

\[P(X=5)=(\begin{array}{c}5\\ 5\end{array})(\frac{2}{3})^{5} (1-\frac{2}{3})^{5-5}\]

\(\displaystyle={0.1317}\)

Thus, the probability that in 30 years all 5 men will be alive is 0.1317.

Step 4

b. The probability that in 30 years at least 3 men will be alive is

\(\displaystyle{P}{\left({X}\geq{3}\right)}={P}{\left({X}={3}\right)}+{P}{\left({X}={4}\right)}+{P}{\left({X}={5}\right)}\)

\[=(\begin{array}{c}5\\ 3\end{array})(\frac{2}{3})^{3} (1-\frac{2}{3})^{5-3}+(\begin{array}{c}5\\ 4\end{array})(\frac{2}{3})^{4} (1-\frac{1}{2})^{5-4}+(\begin{array}{c}5\\ 5\end{array})(\frac{2}{3})^{5} (1-\frac{2}{3})^{5-5}\]

\(\displaystyle={0.3292}+{0.3292}+{0.1317}\)

\(\displaystyle={0.7901}\)

The probability that in 30 years at least 3 men will be alive is 0.7901.

Step 5

c. The probability that in 30 years all only 2 men will be alive is

\[P(X=2)=(\begin{array}{c}5\\ 2\end{array})(\frac{2}{3})^{2} (1-\frac{2}{3})^{5-2}\]

\(\displaystyle={0.1646}\)

Thus, the probability that in 30 years all only 2 men will be alive is 0.1646.