# An insurance salesman sells policies to 5 men the prob.

An insurance salesman sells policies to 5 men the prob. that a mean will be a live in 30 years is $$\displaystyle{\frac{{{2}}}{{{3}}}}$$. Find the prob. that in 30 years: a) all 5 men , b) at least 3 men , c) only 2 men

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ol3i4c5s4hr
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.
###### Not exactly what you’re looking for?
lovagwb
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: $$\displaystyle{P}{\left({x}\geq{3}\right)}={C}{\left({5},{3}\right)}\cdot{\left({\frac{{{2}}}{{{3}}}}\right)}^{{{3}}}\cdot{\left({\frac{{{1}}}{{{3}}}}\right)}^{{{2}}}+{C}{\left({5},{4}\right)}\cdot{\left({\frac{{{2}}}{{{3}}}}\right)}^{{{4}}}\cdot{\left({\frac{{{1}}}{{{3}}}}\right)}^{{{1}}}+{C}{\left({5},{5}\right)}\cdot{\left({\frac{{{2}}}{{{3}}}}\right)}^{{{5}}}={\frac{{{80}}}{{{243}}}}+{\frac{{{80}}}{{{243}}}}+{\frac{{{32}}}{{{243}}}}={\frac{{{192}}}{{{243}}}}$$.
3) Only two men will be alive: $$\displaystyle{P}{\left({x}={2}\right)}={C}{\left({5},{2}\right)}\cdot{\left({\frac{{{2}}}{{{3}}}}\right)}^{{{2}}}\cdot{\left({\frac{{{1}}}{{{3}}}}\right)}^{{{3}}}={\frac{{{40}}}{{{243}}}}$$.
4) At least 1 man will be alive: $$\displaystyle{P}{\left({x}\geq{1}\right)}={1}-{P}{\left({x}={0}\right)}={1}-{C}{\left({5},{0}\right)}\cdot{\left({\frac{{{1}}}{{{3}}}}\right)}^{{{5}}}={\frac{{{242}}}{{{243}}}}$$.
nick1337

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) $$P(x \geq 3) = C(5,3) \cdot (2/3)^{3} \cdot (1/3)^{2} + C(5,4) \cdot (2/3)^{4} \cdot (1/3)^{1} + C(5,5) \cdot (2/3)^{5} = 192/243$$
c) $$P(x=2) = C(5,2) \cdot (2/3)^{2} \cdot (1/3)^{3} = 40/243$$
d) $$P(x \geq 1) = 1 - P(x=0) = 1 - C(5,0) \cdot (1/3)^{5} = 242/243$$