Question

An insurance agent insures lives of $5$ men, all of the same age and in good health. The probability that a man of this age will survive the next $30$ years is known to be $\frac{2}{3}$. Find the probability that in the next $30$ years at most $3$ men will survive

Answer 1

Let $p$ be the probability that a man will survive the next $30$ years.

Then, $p=\frac{2}{3}$
$\therefore q=1-p=1-\frac{2}{3}=\frac{1}{3}$
Given, $n=5$
By Binomial distribution,
$p(x-r){=}^n{C}_r{p}^r{q}^{n-r},0\le r\le n$
$\therefore$ Probability not in the next $30$ years at most $3$ men will survive
$=p(x\le 3)$
$=1-p(x>3)$
$=1-\left[p\right(x=4)+p(x=5\left)\right]$
$=1-\left[{}^5{C}_4{\left(\frac{2}{3}\right)}^4{\left(\frac{1}{3}\right)}^{5-4}{+}^5{C}_5{\left(\frac{2}{3}\right)}^5{\left(\frac{1}{3}\right)}^{5-5}\right]$
$=1-[5{\left(\frac{2}{3}\right)}^4\left(\frac{1}{3}\right)+{\left(\frac{2}{3}\right)}^5]$
$=1-[\frac{80}{243}+\frac{32}{243}]$
$=1-\frac{112}{243}$
$=\frac{131}{243}$