Question

ABC Auto Insurance classifies drivers as good, medium, or poor risks. Drivers who apply to them for insurance fall into these three groups in the proportions 30 percent, 50 percent, and 20 percent, respectively. The probability a "good" driver will have an accident is .01, the probability a "medium" risk driver will have an accident is .03, and the probability a "poor" driver will have an accident is .10. The company sells Mr. Brophy an insurance policy and he has an accident. What is the probability Mr. Brophy is: a. A "good" driver? b. A "medium" risk driver? c. A "poor" driver?

Answer 1

Answer:

a.$P(E_1/A)=0.0789$

b.$P(E_2/A)=0.395$\

c.$P(E_3/A)=0.526$

Step-by-step explanation:

Let $E_1,E_2,E_3$ are the events that denotes the good drive, medium drive and poor risk driver.

$P(E_1)=0.30,P(E_2)=0.50,P(E_3)=0.20$

Let A be the event that denotes an accident.

$P(A/E_1)=0.01$

$P(A/E_2=0.03$

$P(A/E_3)=0.10$

The company sells Mr. Brophyan insurance policy and he has an accident.

a.We have to find the probability Mr.Brophy is a good driver

Bayes theorem,$P(E_i/A)=\frac{P(A/E_i)\cdot P(E_1)}{\sum_{i=1}^{i=n}P(A/E_i)\cdot P(E_i)}$

We have to find $P(E_1/A)$

Using the Bayes theorem

$P(E_1/A)=\frac{P(A/E_1)\cdot P(E_1)}{P(E_1)\cdot P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$

Substitute the values then we get

$P(E_1/A)=\frac{0.30\times 0.01}{0.01\times 0.30+0.50\times 0.03+0.20\times 0.10}$

$P(E_1/A)=0.0789$

b.We have to find the probability Mr.Brophy is a medium driver

$P(E_2/A)=\frac{0.03\times 0.50}{0.038}=0.395$

c.We have to find the probability Mr.Brophy is a poor driver

$P(E_3/A)=\frac{0.20\times 0.10}{0.038}=0.526$