Question

An agent sells life insurance policies to five equally aged, healthy people. According to recent data, the probability of a person living in these conditions for 30 years or more is 2/3. Calculate the probability that after 30 years:

a. All five people are still living.

b. At least three people are still living.

c. Exactly two people are still living.

Answer 1

Let us consider person living in these conditions for $30$ years or more is a success.

Then probability of success is $\frac{2}{3}$ and probability of failure is $\frac{1}{3}$.

$\therefore p=\frac{2}{3}$ and $q=\frac{1}{3}$.

Now we know that for binomial distribution probability of getting $r$ success from $n$ trial is $={^n}C_r.p^r.q^{(n-r)}$

(a) Here $n=5,r=5$

$\therefore$ Probability of all five people are still living is $={^5}C_5.(\frac{2}{3})^5.(\frac{1}{3})^{(5-5)}=(\frac{2}{3})^5$

(b) Here $n=5$ .We have to calculate the probability for $r=3,4,5$ .

$\therefore$ Probability of at least three people are still living is $={^5}C_3.(\frac{2}{3})^3.(\frac{1}{3})^{(5-3)}+{^5}C_4.(\frac{2}{3})^4.(\frac{1}{3})^{(5-4)}+{^5}C_5.(\frac{2}{3})^5.(\frac{1}{3})^{(5-5)}$

$=24.(\frac{2^3}{3^5})$

(c) Here $n=5,r=2$

Probability of exactly two people are still living $={^5}C_2.(\frac{2}{3})^2.(\frac{1}{3})^{(5-2)}$ $={^5}C_5.(\frac{2}{3})^2.(\frac{1}{3})^{3}$ $=\frac{4}{3^5}$