An insurance salesman sells policies to 5 men the prob. that a mean will be...

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 ${\displaystyle \frac{2}{3}\text{and}n=5}$.
Here, men are independent and probability of success is constant. Hence, X follows binomial distribution with parameters ${\displaystyle n=5\text{and}p=\frac{2}{3}}$.
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
${\displaystyle P(X\ge 3)=P(X=3)+P(X=4)+P(X=5)}$
$$=(\begin{array}{c}5\\ 3\end{array})(\frac{2}{3}{)}^3(1-\frac{2}{3}{)}^{5-3}+(\begin{array}{c}5\\ 4\end{array})(\frac{2}{3}{)}^4(1-\frac{1}{2}{)}^{5-4}+(\begin{array}{c}5\\ 5\end{array})(\frac{2}{3}{)}^5(1-\frac{2}{3}{)}^{5-5}$$
${\displaystyle =0.3292+0.3292+0.1317}$
${\displaystyle =0.7901}$
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.

1) Probability that all men will be alive: ${\displaystyle P={\left(\frac{2}{3}\right)}^5=\frac{32}{243}}$.
2) Probability that at least 3 men will be alive: ${\displaystyle P(x\ge 3)=C(5,3)\cdot {\left(\frac{2}{3}\right)}^3\cdot {\left(\frac{1}{3}\right)}^2+C(5,4)\cdot {\left(\frac{2}{3}\right)}^4\cdot {\left(\frac{1}{3}\right)}^1+C(5,5)\cdot {\left(\frac{2}{3}\right)}^5=\frac{80}{243}+\frac{80}{243}+\frac{32}{243}=\frac{192}{243}}$.
3) Only two men will be alive: ${\displaystyle P(x=2)=C(5,2)\cdot {\left(\frac{2}{3}\right)}^2\cdot {\left(\frac{1}{3}\right)}^3=\frac{40}{243}}$.
4) At least 1 man will be alive: ${\displaystyle P(x\ge 1)=1-P(x=0)=1-C(5,0)\cdot {\left(\frac{1}{3}\right)}^5=\frac{242}{243}}$.

Step-by-step explanation:
$n=5$
$p=2/3$
$q=1/3$
$P(x)=C(n,r)\cdot p^x\cdot q^{n-x}$
a) $P(x=5)=C(5,5)\cdot (2/3{)}^5=32/243$
b) $P(x\ge 3)=C(5,3)\cdot (2/3{)}^3\cdot (1/3{)}^2+C(5,4)\cdot (2/3{)}^4\cdot (1/3{)}^1+C(5,5)\cdot (2/3{)}^5=192/243$
c) $P(x=2)=C(5,2)\cdot (2/3{)}^2\cdot (1/3{)}^3=40/243$
d) $P(x\ge 1)=1-P(x=0)=1-C(5,0)\cdot (1/3{)}^5=242/243$