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Suppose you have a particularly busy intersection. That intersection averages about 4.3 accidents per week. Now, that's not a whole lot of accidents per week for a busy intersection with a lot of cars passing through it. The probability of having an accident at that intersection is fairly small. All other things being equal, what's the probability that the intersection experiences a week with just one accident?
This distribution can be solved using something called the Poisson distribution. It's a distribution that works well for rare events--rare meaning the probability of success is very low, but the number of trials is very high. Out of a very large number of trails, you end up with only a few successes.
The Poisson distribution is nice because it doesn't require the use of n and p--all you need is the typical rate of occurrence and the number of events that you are anticipating to occur during this time frame. The following formula demonstrates this relationship:
If the average number of successes in a given time frame is the Greek letter lambda and x is the number of potential successes that you could have, the probability that you get exactly k successes is equal to lambda to the k--the expected number to the number of successes--times the number e to the negative lambda divided by k factorial.
So, k factorial (k followed by the exclamation point) means that you start from k and multiply by k minus 1, k minus 2, all the way down to 1. That means that 4 factorial would be 4 times 3 times 2 times 1.
Going back to our busy intersection example, lambda--the expected rate of occurrence--was 4.3. The number of successes that you wanted this week was 1.
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