1. ## Conditional probability.

An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C. The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%. The probabilities that a driver in one of these classes will have an accdient within one year are given by 0.01, 0.02, and 0.03 respectively. After purchasing an insurance policy, a driver has an accident within the first year. What is the probability that the driver is of class A?

I think that it's 1/6, because you'd do 0.01/0.06, since you already know that the driver has had the accident, but I'm not too sure. One other possibility that I'm considering would be 0.2 * 0.01, but I feel like this doesnt account for the fact that we are assuming that the driver is already in an accident.

2. We are trying to figure out: (prob of driver in class A having an accident)/(prob of any driver having an accident). So we know the probability of some driver in class A, B, or C having an accident within the first year: (prob of being in class * prob of having accident) for each class. Sum those up to find the prob of any driver and you should be on the right track. Just remember when something is given that is the new denominator in the probability because you are only look at those people because it is already given that this event has happened.

Does this make sense? I got scolded the other day for giving too much of an answer away, so I am trying to help you without being to obvious or giving you the answer. Sorry if it came across as confusing.

3. Use Bayes' Theorem. Post if you have any problems.

4. Originally Posted by tommiegirl13
We are trying to figure out: (prob of driver in class A having an accident)/(prob of any driver having an accident).
No you are not, as someone else has already posted, use Bayes theorem.

CB