Poisson Distribution Problem Exam P

Question

I am prepping for Exam P and getting a little confused on how to solve/set up part B?:

Claims filed in a year by a policyholder of an insurance company have a Poisson distribution with $\lambda=.40$. The number of claims filed by two different policyholders are independent events.

(a) If two policyholders are selected at random, what is the probability that each of them will file one claim during the year? (b) What is the probability that at least one of them will file no claims?

a). $[\exp(-0.4) * 0.4^1]/1! = \text{Probability 1 claim is filed} \implies$ $P(A \cap B) = 0.268128 * 0.268128 = 0.0719$

b). I'm thinking the answer is $[\exp(-0.4) * 0.4^0]/0! + [\exp(-0.4) * 0.4^1]/1! = 0.938448064$

But the book says it's $.8913$. What am I doing wrong?

Answer 1

Let $A$ be the event that person A files at least one claim and $B$ be the event that person B files at least one claim. Note that $$ P(\bar{A}\cup\bar{B}) =1-P(A\cap B) =1-P(A)P(B) =1-(1-e^{-0.4})^2 =.8913 $$ where we have used independence in the second equality.

Answer 2

For the second part of the question, you may partition all of the outcomes into the following mutually exclusive events:

1. At least one of the two policyholders files no claims.
2. Both policyholders file at least one claim.

Exactly one of these two events must occur. Therefore, the sum of the probabilities of these two events equals $1$; that is to say, they are complementary events, and if we can compute the probability of either one, we can immediately find the probability of the other event since the total probability adds to $1$.

So, let's look at the second event. This in itself is the intersection of two (independent) events; that is to say, if both policyholders file at least one claim, that means policyholder 1 files at least one claim, and policyholder 2 files at least one claim.

So let's look at the probability that a given policyholder files at least one claim. This is itself the complement of the probability that a given policyholder files no claim; i.e., if $X_1$ is the number of claims filed by policyholder 1 in a year, then $$\Pr[X_1 \ge 1] = 1 - \Pr[X_1 = 0] = 1 - e^{-\lambda} \frac{\lambda^0}{0!} = 1 - e^{-\lambda} = 1 - e^{-0.4}.$$ But the behavior of policyholder 2 is independent and identically distributed, thus the same is true of policyholder 2: $$\Pr[X_2 \ge 1] = 1 - e^{-0.4}.$$ And since their outcomes are independent, the probability of both policyholders having no claims is $$\Pr[(X_1 \ge 1) \cap (X_2 \ge 1)] \overset{\text{ind}}{=} \Pr[X_1 \ge 1]\Pr[X_2 \ge 1] = (1 - e^{-0.4})^2.$$ But remember, this is event 2 in the list at the top of this post. The desired event is event 1, the complement. So the answer is $$\Pr[(X_1 = 0) \cup (X_2 = 0)] = 1 - \Pr[(X_1 \ge 1) \cap (X_2 \ge 1)] = 1 - (1 - e^{-0.4})^2.$$