An insurance company believes that people can be divided into two classes: those who are accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at some time within a fixed 1-year period with probability .4, whereas this probability decreases to .2 for a person who is not accident prone. If we assume that 30 percent of the population is accident prone, what is the probability that someone has an accident in the second year given that he or she had no accidents in the first year?

My attempt. Let $A_1$ and $A_2$ denote the events that an accident occured in year 1 and 2 respectively. Let $p$ denote the event that one is accident prone.

$$P(A_2\mid A_1^c)=\frac{P(A_2\cap A_1^c)}{P(A_1^c)}=\frac{P(A_2\cap A_1^c)}{P(p)P(A_1^c\mid p)+P(p^c)P(A_1^c\mid p^c)}=\frac{P(A_2\cap A_1^c)}{0.3\times 0.6 + 0.7\times 0.8}=\frac{P(A_2\cap A_1^c)}{0.74}$$

I just don't know what to do with $P(A_2\cap A_1^c)$. I could express it as ${P(A_2)P(A_1^c\mid A_2)}$ but I don't know what to do with it. Some help would be greatly appreciated.

  • 1
    $\begingroup$ You've nearly got there. Just separate $P(A_2 \cap A_1^c)$ according to $p$ and $p^c$ as you did with the denominator. You haven't specified any other dependencies, so the probabilities will multiply. $\endgroup$
    – user383778
    Commented Dec 2, 2016 at 20:25

1 Answer 1


Keeping your notation, although capital letters for events (like your p) usually makes things a bit clearer, the only term you have left to evaluate is $$ P(A_2 \cap A_1^c ) = p(A_2 \cap A_1^c | p)P(p) + p(A_2 \cap A_1^c | p^c )P(p^c), $$ and these terms should be easy to evaluate $$ P(A_2 \cap A_1^c |p) = P(A_2 | A_1^c,p) P(A_1^c |p )=P(A_2|A_1^c,p)(1-P(A_1|p)) $$ and then finally without any other specified information it would suggest that the probability of an accident in the second fixed period given no accident is just the probability of an accident in a fixed year period $$ P(A_2 |A_1^c,p) = P(A_2 | p) $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .