1. ## Conditional probability problem

An insurance company believes that drivers can be divided into two classes, high-risk and low-risk. According to past data, a high-risk driver has an accident with probability 0.3 during a typical year. On the other hand, a low-risk driver has an accident with probability 0.1 during a typical year. Furthermore, 20% of policyholders are high-risk.

(a) What is the probability that a new policyholder will have an accident within a year of purchasing a policy?

(b) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that that driver is high-risk?

(c) Suppose that a new policyholder has an accident in year one. What is the probability that the policyholder will have an accident in year two? (You may assume that for a given individual, the occurrence of an accident in one year is independent of the occurrence of an accident in another year.)

Using A = "having and accident" and H = "high-risk driver", I found part (a) to be P(A) = 0.14 and part (b) to be P(H|A) = 0.4286 without any problem, but I don't know how to set up part (c). I know I need to take into account that, since the driver had an accident the first year, he may be high-risk. Right now, I'm thinking it could be P(A|H)P(H|A) + P(A|H')P(H'|A), but I'm not sure.

2. Originally Posted by uberbandgeek6
An insurance company believes that drivers can be divided into two classes, high-risk and low-risk. According to past data, a high-risk driver has an accident with probability 0.3 during a typical year. On the other hand, a low-risk driver has an accident with probability 0.1 during a typical year. Furthermore, 20% of policyholders are high-risk.

(a) What is the probability that a new policyholder will have an accident within a year of purchasing a policy?

(b) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that that driver is high-risk?

(c) Suppose that a new policyholder has an accident in year one. What is the probability that the policyholder will have an accident in year two? (You may assume that for a given individual, the occurrence of an accident in one year is independent of the occurrence of an accident in another year.)

Using A = "having and accident" and H = "high-risk driver", I found part (a) to be P(A) = 0.14 and part (b) to be P(H|A) = 0.4286 without any problem, but I don't know how to set up part (c). I know I need to take into account that, since the driver had an accident the first year, he may be high-risk. Right now, I'm thinking it could be P(A|H)P(H|A) + P(A|H')P(H'|A), but I'm not sure.
For c you can use the half-Bayes' theorem:

$\displaystyle P(U \wedge V)=P(U|V)P(V)$

Here $\displaystyle $$U denotes an accident in year 2 and \displaystyle$$ V$ an accident in year 1.

$\displaystyle P(V)=P(H)\times0.3+(1-P(H))\times 0.1$

and:

$\displaystyle P(U \wedge V)=P(H)(0.3)^2+(1-P(H))(0.1)^2$

etc

CB

3. I don't follow how you went from P(U|V)P(V) to P(H)P(0.3)^2 + (1 - P(H))(0.1)^2.

4. Originally Posted by uberbandgeek6
I don't follow how you went from P(U|V)P(V) to P(H)P(0.3)^2 + (1 - P(H))(0.1)^2.
I didn't, I gave an equation containing three probabilities and showed you how to compute two of them leaving you to compute the third P(U|V) from them.

P(H)P(0.3)^2 + (1 - P(H))(0.1)^2 is the apriori probability of a driver having acidents in year 1 and in year 2 that is P(U and V).

CB

5. Okay, so I'm really solving for P(U|V), which makes a lot more sense. Thank you!