Question

An insurance sales representative sells policies to 5 men, all identical age and in good health.
According to actual 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
(a) all 5 men; (b) at least 3 men; (c) none will be alive

Answer 1

Let $X=$ the number of men, who will be alive in 30 years: $X\sim B(n, p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given that $n=5, p={2 \over 3}$

(a)

$$P(X=5)=\binom{5}{5}({2 \over 3})^5(1-{2 \over 3})^{5-5}={32 \over 243}\approx0.1317$$

(b)

$$P(X\geq3)=P(X=3)+P(X=4)+P(X=5)=$$

$$=\binom{5}{3}({2 \over 3})^3(1-{2 \over 3})^{5-3}+\binom{5}{4}({2 \over 3})^4(1-{2 \over 3})^{5-4}+$$

$$+\binom{5}{5}({2 \over 3})^5(1-{2 \over 3})^{5-5}={80 \over 243}+{80 \over 243}+{32 \over 243}=$$

$$={64 \over 81}\approx0.7901$$

(c)

$$P(X=0)=\binom{5}{0}({2 \over 3})^0(1-{2 \over 3})^{5-0}={1 \over 243}\approx0.0041$$