Zerrilloh6

2021-12-19

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

ol3i4c5s4hr

Expert

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 .
Here, men are independent and probability of success is constant. Hence, X follows binomial distribution with parameters .
The probability mass function of binomial random variable X is
$P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}\left(1-p{\right)}^{n-x};x=0,1,...,n$
Step 3
a. The probability that in 30 years all 5 men will be alive is
$P\left(X=5\right)=\left(\begin{array}{c}5\\ 5\end{array}\right)\left(\frac{2}{3}{\right)}^{5}\left(1-\frac{2}{3}{\right)}^{5-5}$
$=0.1317$
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
$P\left(X\ge 3\right)=P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)$
$=\left(\begin{array}{c}5\\ 3\end{array}\right)\left(\frac{2}{3}{\right)}^{3}\left(1-\frac{2}{3}{\right)}^{5-3}+\left(\begin{array}{c}5\\ 4\end{array}\right)\left(\frac{2}{3}{\right)}^{4}\left(1-\frac{1}{2}{\right)}^{5-4}+\left(\begin{array}{c}5\\ 5\end{array}\right)\left(\frac{2}{3}{\right)}^{5}\left(1-\frac{2}{3}{\right)}^{5-5}$
$=0.3292+0.3292+0.1317$
$=0.7901$
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
$P\left(X=2\right)=\left(\begin{array}{c}5\\ 2\end{array}\right)\left(\frac{2}{3}{\right)}^{2}\left(1-\frac{2}{3}{\right)}^{5-2}$
$=0.1646$
Thus, the probability that in 30 years all only 2 men will be alive is 0.1646.

lovagwb

Expert

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

nick1337

Expert

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

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