This question is #63 in Chapter 2 of Modern Mathematical Statistics with Applications by Jay Devore and Kenneth Berk (Page 84, 2nd edition).
A professional organization sells term life insurance and medical insurance.
- Of those who have just life insurance, 70% will renew next year
- Of those who have just medical insurance, 80% will renew next year
- Of those who have both types of insurance, 90% will renew at least one of them next year.
Of the policy holders
- 75% have term life insurance
- 45% have major medical
- 20% have both
a. Calculate the percentage of policyholders that will renew at least one policy next year.
b. If a randomly selected policy holder does in fact renew next year, what is the probability that he or she has both life and major medical insurance?
Here are my attempts to answer the 2 questions. Do they look right to you?
Let
- $L = \text{have life insurance}$
- $M = \text{have medical insurance}$
- $R = \text{will renew the insurance policy next year}$
a) $P(R) = P(R \cap L \cap M^c) + P(R \cap L^c \cap M) + P(R \cap L \cap M) \\ = P(R | L \cap M^c) * P(L \cap M^c) + P(R | L^c \cap M) * P(L^c \cap M) + P(R | L \cap M) * P(L \cap M)$
$P(L \cap M^c) = P(L) - P(L \cap M) = 0.75 - 0.20 = 0.55$ $P(L^c \cap M) = P(M) - P(L \cap M) = 0.45 - 0.20 = 0.25$
$P(R) = 0.70*0.55 + 0.80*0.25 + 0.90*0.20 = 0.765$
b) $P(L \cap M | R) = P(L \cap M \cap R) \ \div \ P(R) = 0.90 * 0.20 \ \div \ 0.765 = 0.235$