Question

6. If 10P30-0.8 and 20Ps0-0.6, find the probability that of three persons now aged 30, at least two will die between ages 40 and 50.] 0.104

Answer 1

Let T be future lifetime random variable, then

₁₀p₃₀ = P(T > 40) = 0.8

₂₀p₃₀ = P(T > 50) = 0.6

Probability that the person at age 30 will die between ages 40 and 50 = P(T > 40) - P(T > 50) = 0.8 - 0.6 = 0.2

Number of people (X) die out of 3 persons will follow binomial distribution with parameters n = 3, p = 0.2

By Binomial distribution, probability that at lest two will die between ages 40 and 50 is,

P(X $\ge$ 2) = P(X = 2) + P(X = 3)

= ³C₂ * 0.2² * (1 - 0.2)¹ +  ³C₃ * 0.2³ * (1 - 0.2)⁰

= 3 * 0.04 * 0.8 + 0.008

= 0.104