• Aug 6th 2006, 10:05 PM
bobby77
An automobile insurance company classifies drivers as Class A(good risks), class B(medium risks), and class C(poor risks). They believe that class A risks constitute 30% of the drivers who apply to them for insurance, class B 50%, and class C(20%). The probability that a class A driver will have one or more accidents in any 12 month period is .01, for class B is .03, and for class C is .10. The company sells Mr. Jones an insurance policy and within a year he has an accident. What is the probability that Mr. Jones is a class B risk?
• Aug 7th 2006, 04:05 AM
galactus
Are you familiar with Bayes' theorem?.

Let's let D=portion of drivers from the respective classes

$\displaystyle P(B|D)=\frac{P(D|B)P(B)}{P(D|B)+P(D|A)P(A)+P(D|C)P (C)}$

$\displaystyle \frac{(.5)(.03)}{(.5)(.03)+(.30)(.01)+(.20)(.10)}= .395$

The probability the driver is class B is about 39.5%.

Check me out here. It's easy to get mixed up. It's been awhile since I used Bayes' theorem.
• Aug 7th 2006, 05:27 AM
Soroban
Hello, bobby77!

Galactus' answer is absolutely correct . . .

Quote:

An automobile insurance company classifies drivers as:
. . Class A (good risks), class B (medium risks), and class C (poor risks).
They found that class A constitutes 30% of their drivers, class B 50%, and class C 20%.
The probability that a driver will have one or more accidents in any 12-month period is:
. . 0.01 for class A, 0.03 for class B, and 0.10 for class C.

The company sells Mr. Jones an insurance policy and within a year he has an accident.
What is the probability that Mr. Jones is a class B risk?

I checked the answer with a baby-talk approach.
. . (You can use this when you're desperate.)

Suppose the insurance company has exactly 1000 customers.

Then 30% (300) are class A, 50% (500) are class B, and 20% (200) are class C.

Of the 300 class A drivers, 1% will have an accident . . . $\displaystyle 3$

Of the 500 class B drivers, 3% will have an accident . . . $\displaystyle 15$

Of the 200 class C drivers, 10% will have an accident . . . $\displaystyle 20$

Hence, $\displaystyle 3 + 15 + 20 = 38$ drivers will have accidents.

Mr. Jones had an accident, so he is one of those 38 drivers
. . . and 15 of them are class B drivers.

Therefore, the probability that Mr. Jones is class B is:
. . $\displaystyle \frac{15}{38} \:=\:0.394736842 \:\approx\:39.5\%$
. . . ta-DAA!

Nice work, Cody!

• Aug 7th 2006, 05:57 AM
galactus
Thanks for the confirmation, Soroban. I am glad I remembered Bayes' theorem.