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An insurance company has 50% urban and 50% rural customers. If every year each urban customer has an accident with probability μ and each rural customer has an accident with probability λ. Assume that the chance of an accident is independent from year to year and from customer to costumer. This is another way to say, conditioned on being and urban or rural the chance of having an accident each year is independent.

A costumer is randomly chosen. Let An be the chance this customer has an accident in year n. Let U denote the event that this costumer is urban and R the event that the customer is rural.


  1. Find $P(A2|A1)$.

  2. Are A1 and A2 independent in general ? Are there any conditions when it is true if not in general ?

  3. Show that P(A2|A1)≥P(A2).

To answer this question it is useful to know that for any positive a and b, one has $(a+b)^2<2(a^2+b^2)$ as long as a≠b. In the case $a≠b$, one has of course $(a+b)^2=2(a^2+b^2)$. To prove this inequality, first show that $(a+b)^2+(a−b)^2=2(a^2+b^2)$ and then use that fact that $(a−b)^2>0$.

  1. Find the probability that a driver has an accident in the 3rd year given that they had one in the 1st and 2nd year.

  2. Find the probability that a driver has an accident in the n-th year given that they had one in all of the previous years. What is the limit as n→∞ ?

  3. Find the probability that a diver is a urban diver given that they had an accident in two successive years.


I spent about 30 minutes doing this last night without getting anywhere. I think I am either reading the question wrong or the question is written wrong.

  1. P(A1|A2) = P(A2)P(A1) = (0.5$\mu$ + 0.5$\lambda$)$^2$.

  2. Not really sure how to interpret this.

  3. This is where my confusion kicks in. In #1, I got that P(A2|A1) = (0.5$\mu$ + 0.5$\lambda$)$^2$. Looking at the hint, I thought that this would mean that P(A2|A1)

Any help is appreciated! Thank you in advance.

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It helps to first establish the given probabilities of the events described in the question.

First, we have $\Pr[U] = \Pr[R] = 1/2$; the probability that a randomly selected customer is urban is equal to the probability that a randomly selected customer is rural, and both are equal to $1/2$.

We are also given that $\Pr[A_n \mid U] = \mu$ and $\Pr[A_n \mid R] = \lambda$; that is, the probability of a given urban customer having an accident in year $n$ is $\mu$, and the probability of a given rural customer having an accident in year $n$ is $\lambda$.

We are asked to find $\Pr[A_2 \mid A_1]$. To do this, we must use Bayes' rule: $$\Pr[X \mid Y] = \frac{\Pr[Y \mid X]\Pr[X]}{\Pr[Y]},$$ for two events $X$ and $Y$. Applying this with the choice $X = U$, $Y = A_1$, we get $$\Pr[U \mid A_1] = \frac{\Pr[A_1 \mid U]\Pr[U]}{\Pr[A_1]} = \frac{\mu/2}{\Pr[A_1]}.$$ But what is the unconditional probability $\Pr[A_1]$, that is, what is the chance that a randomly selected customer has an accident in the first year? Clearly, it relates to all four probabilities we were given--the probability of choosing an urban vs. rural customer; and the probabilities of customers of each type having an accident. The answer is to use the law of total probability: $$\Pr[X] = \Pr[X \mid Y_1]\Pr[Y_1] + \Pr[X \mid Y_2]\Pr[Y_2] + \cdots + \Pr[X \mid Y_n]\Pr[Y_n],$$ for an event $X$ and for a set of mutually disjoint events $Y_1, Y_2, \ldots, Y_n$ that satisfy $\Pr[Y_1] + \Pr[Y_2] + \cdots + \Pr[Y_n] = 1$. This is the discrete, finite version of the law. Applied to this case, we see that $$\Pr[A_1] = \Pr[A_1 \mid U]\Pr[U] + \Pr[A_1 \mid R]\Pr[R] = \frac{\mu}{2} + \frac{\lambda}{2} = \frac{\mu+\lambda}{2}.$$ Therefore, $$\Pr[U \mid A_1] = \frac{\mu}{\mu+\lambda}.$$ And similarly, $$\Pr[R \mid A_1] = \frac{\lambda}{\mu+\lambda}.$$ Now to find $\Pr[A_2 \mid A_1]$, it serves to condition $A_2$ on the customer type: $$\Pr[A_2 \mid A_1] = \Pr[A_2 \mid U]\Pr[U \mid A_1] + \Pr[A_2 \mid R]\Pr[R \mid A_1] = \frac{\mu^2+\lambda^2}{\mu+\lambda}.$$ To understand why this works, the LHS says that the customer had an accident in the first year. This customer is urban with probability $\Pr[U \mid A_1] = \mu/(\mu+\lambda)$, and if so, will have another accident next year with probability $\Pr[A_2 \mid U] = \mu$. And if this customer is rural with probability $\Pr[R \mid A_1] = \lambda/(\mu+\lambda)$, they will have another accident with probability $\Pr[A_2 \mid R] = \lambda$. This is the answer to (1).

To answer (2), independence is true if $\Pr[A_2 \mid A_1] = \Pr[A_2]$; that is, having observed $A_1$ does not change the probability of observing $A_2$. Is this true? You should argue that $\Pr[A_2] = \Pr[A_1]$: why?

For part (3), use what you found in parts (1) and (2), and the hint; now that you know the correct probabilities, the hint should make sense.

For parts (4-6), you need to extend your reasoning in the previous parts. The idea is to reason that if a randomly selected driver has accidents in years $1, 2, \ldots, n-1$, then call this event $B_{n-1} = A_1 \cap A_2 \cap \ldots \cap A_{n-1}$. Then use the law of total probability and Bayes' rule again to write $$\Pr[U \mid B_{n-1}] = \frac{\Pr[B_{n-1} \mid U]\Pr[U]}{\Pr[B_{n-1} \mid U]\Pr[U] + \Pr[B_{n-1} \mid R]\Pr[R]}$$ where I have combined the two formulas into a single expression. The only difference now is that you need to know the probabilities of a driver of each type having $n-1$ consecutive accidents, rather than just a single accident.

Then repeat the above for $\Pr[R \mid B_{n-1}]$, which should be easy to see in advance; finally, follow the method at the end of part (1) to reason that $$\Pr[A_n \mid B_{n-1}] = \ldots?$$ I'm leaving this to you.


As an instructive exercise, once you finish this entire question, go back and redo everything in the generalized case where $\Pr[U] = p$, $\Pr[R] = q = 1-p$, where $0 \le p, q \le 1$.

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  • $\begingroup$ Thank you for the detailed explanation. I actually figured out most of this problem before you posted the solution but seeing your clean logic is just awe-striking. Again - thank you! $\endgroup$ – user3642365 Jan 23 '15 at 20:10

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