The probability that a person of a particular age will be alive 30 years hence is 0.25. An Insurance Policy is bought by 6 persons of identical age and health conditions. Find the probability that 30 years hence, of these 6 persons, what is the likely estimation that i) Exactly three persons will be alive, ii) No person will be alive.

Let $X$ denote the random variable that a person will be alive hence. Then $X\sim Binomial(n=6,p=0.25)$ given as,

$p(X=x)=\binom{6}{x}0.25^x0.75^{6-x},\space x=0,1,2,3,4,5,6$

$a)$

The probability that exactly 3 persons will be alive is,

$p(X=3)=\binom{6}{3}0.25^30.75^{3}=20\times0.015625\times0.421875=0.1318$

The probability that exactly 3 persons will be alive is 0.1318.

$b)$

We determine the probability,

$p(X=0)=\binom{6}{0}0.25^00.75^{6}=0.178$

Therefore, the probability that no person will be alive is 0.178.