Question

An insurance salesman sells policies to 5 men, all of identical age in good health. According to

the actuarial tables the probability that a man of this particular age will be alive 30 years hence is

2 /3 . Find the probability that in 30 years

(i) All men will be alive

(ii) At least 3 men will be alive

(iii) Only two men will be alive

(iv) At least 1 man will be alive

Answer 1

$n=5 \\ p = \frac{2}{3} \\ q = \frac{1}{3} \\ P(X=x)=C(n,x)p^x q^{n-x}$

(i) All men will be alive

$P(X=5) = \binom{5}{5}(\frac{2}{3})^5(\frac{1}{3})^{5-5} = 0.1317$

(ii) At least 3 men will be alive

$P(X≥3) = P(X=3) + P(X=4) + P(X=5) \\ = \binom{5}{3}(\frac{2}{3})^3(\frac{1}{3})^{5-3} + \binom{5}{4}(\frac{2}{3})^4(\frac{1}{3})^{5-4} + \binom{5}{5}(\frac{2}{3})^5(\frac{1}{3})^{5-5} \\ = 0.790123$

(iii) Only two men will be alive

$P(X=2) = \binom{5}{2}(\frac{2}{3})^2(\frac{1}{3})^{5-2} = 0.164609$

(iv) At least 1 man will be alive

$P(X≥1) = 1 - P(X<1) \\ = 1 -P(X=0) \\ = 1 -[\binom{5}{0}(\frac{2}{3})^0(\frac{1}{3})^{5-0}] \\ = 1 -0.004115 \\ = 0.995885$