An agent sells life insurance policies to twelve equally aged, healthy people. According to recent data, the probability of a person living in their conditions for 30 years or more is 2/5. Calculate the probability that after 30 years:

(a)0.1317

(b)0.1646

(c)0.7901

Step-by-step explanation:

This is a Binomial distribution problem.

For X ∼ Bin(n, p),

(a)All five people are still living.

The number of people still living has a Binomial(n = 5, p = 2/3)

=

≈0.1317

(b)Exactly two people are still living

≈0.1646

(c) at least three people are still living.

P(X≥3)=P(X=3)+P(X=4)+P(X=5)

=0.3292

=0.3292

≈0.1317

P(X≥3)=0.3292+0.3292+0.1317

≈0.7901