Thread: Trouble with Probability

Can someone help with the following:

For simplicity, let us assume that there are two kinds of drivers. The safe drivers, who are 70% of the population, have probability 0.1 of causing an accident in a year. The rest of the population are accident makers, who have probability 0.5 of causing an accident in a year. The insurance premium is $400 times one's probability of causing an accident in the following year. A new subscriber has an accident during the first year. What should be his insurance premium for the next year?

Thanks for any help.

Originally Posted by JimmyT

Can someone help with the following:

For simplicity, let us assume that there are two kinds of drivers. The safe drivers, who are 70% of the population, have probability 0.1 of causing an accident in a year. The rest of the population are accident makers, who have probability 0.5 of causing an accident in a year. The insurance premium is $400 times one's probability of causing an accident in the following year. A new subscriber has an accident during the first year. What should be his insurance premium for the next year?

Thanks for any help.

This is an excercise in using Bayes' theorem.

The prior probability that a driver is safe is p_0(safe) = 0.7, and the
probability of an accident:

p_a0=0.7*0.1+0.3*0.5 = 0.22

Then:

p(safe|accident) = p_0(safe)p(accident|safe)/p_a0 = 0.7*0.1/0.22 ~= 0.3182

So his probability of an accident in the next year is:

0.3182*0.1+(1-0.3182)*0.5 ~= 0.3727,

so the premium should be 400*0.3727 = $149.09

RonL