Question

Q.2 An insurance salesman sells polices to 6 men, all of identical age and in good health. According to the actuarial tables, the probabilities that a man of this particular age will be alive 30 years hence is . Find the probability that in 30 years (i) all men (ii) at least 4 men, (iii) only three men, (iv) at most one man will be alive.

Answer 1

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

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

Given $n=6, p=2/3$

(i)

$$P(X=6)=\dbinom{6}{6}({2\over 3})^6(1-{2\over3 })^{6-6}=\dfrac{64}{729}$$

(ii)

$$P(X\geq4)=P(X=4)+P(X=5)+P(X=6)=$$

$$=\dbinom{6}{4}({2\over 3})^4(1-{2\over3 })^{6-4}+\dbinom{6}{5}({2\over 3})^5(1-{2\over3 })^{6-5}+$$

$$+\dbinom{6}{6}({2\over 3})^6(1-{2\over3 })^{6-6}=15(\dfrac{16}{729})+6(\dfrac{32}{729})+$$

$$+\dfrac{64}{729}=\dfrac{496}{729}$$

(iii)

$$P(X=3)=\dbinom{6}{3}({2\over 3})^3(1-{2\over3 })^{6-3}=\dfrac{160}{729}$$

(iv)

$$P(X\leq1)=P(X=0)+P(X=1)=$$

$$=\dbinom{6}{0}({2\over 3})^0(1-{2\over3 })^{6-0}+\dbinom{6}{1}({2\over 3})^1(1-{2\over3 })^{6-1}=$$

$$=\dfrac{1}{729}+\dfrac{12}{729}=\dfrac{13}{729}$$