This just in.... Midterm has craaaaaaaazy insurance claim problem!

I'll tell you what I got. I can't promise that it's right, but if you think I got it wrong, post a comment or shoot me an email. I'm doing this from memory, so if I'm not remembering the question correctly in some substantial way, let me know for sure.

Q. An insurance company determines that they receive an average on nine claims in an average week. Claims in any given week do not effect the chances of claims in other weeks.

For a given week:

A. What is the probability of receiving 7 claims.

B. Given that there was a claim, what is the probability that there were 7 claims.

Answer: Ok, this is clearly an application of the Poisson distribution.

For part A, just plug the numbers into the formula and you get

f(x=7) = (e^-9)(9^7)/7! = (0.0001234)(4782969)/5040 = 0.1171 = 11.71%

BTW, I found a handy-dandy Poisson distribution probability calculator online at Stattrek. I plugged in our numbers there and it calculated the same answer. Ok, that was the easy part. Now on to part B...

Part B tells us that there was at least one claim. We can calculate P(x>=1) = 1 - P(x=0). And P(x=0) = (e^-9)(9^0)/0! = e^-9 = 0.0001234. This makes sense intuitively because the probability of there being no claims at all when the average is 9 is pretty slim. Really slim, in fact.

So, P(x>=1) = 1-0.0001234 = 0.9998766. And we're being asked to determine P(x=7 | x>=1).

By definition of conditional probability:

P(x=7 | x>=1) = P( x=7 and x>=1) / P(x>=1)

We know the denominator. We calculated it to be 0.9998766.

To calculate the numerator (and though we really don't need to calculate anything, we'll try to be rigorous), we note that P(x=7) is clearly mutually exclusive from P(x=0) and therefore P(x=7 and x=0) = 0. And since x=0 and x>=0 are exhaustive of the sample space, we therefore know that P(x=7 and x>=0) = P(x=7). That's our numerator!

So, P(x=7 | x>=1) = P( x=7) / P(x>=1) = 0.1171 / 0.9998766 = 0.1171145 = 11.71145%

This is only slightly more than the probability in part A, but that makes sense since the condition in part B that there was at least one claim didn't significantly decrease the sample space.

That was the only really tricky question on the midterm, imho, although there were plenty of places that one could make a careless mistake. I noticed that most of the incorrect multiple choice answers were offered up to entice you if you made a quick mistake and didn't check your work.

Slightly tricky: Two mutually exclusive events are not independent. This is evident when you look at the test for independence:

P(A and B ) = P(A)P(B)

Well, if they events are mutually exclusive then P(A and B) = 0. Therefore it's impossible for them to be independent if both P(A) and P(B) are non-zero.

Answer to the bonus question: Snow and cold weather are independent since P(snow and cold) = P(snow)P(cold)

## Thursday, January 31, 2008

### The Midterm

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