# Here is the question I'm struggling with: A small city has 2 ambulances. Emergency calls for the ambulances arrive randomly with an average of 0.2 calls per hour. They are concerned about the...

Here is the question I'm struggling with:

A small city has 2 ambulances. Emergency calls for the ambulances arrive randomly with an average of 0.2 calls per hour. They are concerned about the possibility of both ambulances being busy when an additional call comes in. What is the probability of more than 2 calls in a 1-hour period? Determine the correct distribution, explain why it is the best distribution to use, and find the probability.

I've been using a Poisson distribution.

### 1 Answer | Add Yours

Using the Poisson distribution is correct -- the independent variable is occuring over a period of time.

`P(X,lambda)=(e^(-lambda)lambda^X)/(X!)` where X=0,1,2,... is the number of occurences, `lambda` is the mean number of occurences over time.

In order to find the probability of `X>2` we take `1-[P(0,lambda)+P(1,lambda)+P(2,lambda)]` -- this is the complement of the probability that `X<=2` .

By hand:

`P(0,.2)=(e^(-.2)(.2)^0)/(0!)=e^(-.2)~~.8187`

`P(1,.2)=(e^(-.2)(.2)^1)/(1!)=.2e^(-.2)~~.1637`

`P(2,.2)=(e^(-.2)(.2)^2)/(2!)=(.04e^(-.2))/2~~.0164`

So the probability of getting no calls in some hour is about 82%, the probability of getting 1 call is about 16%, and the probability of getting 2 calls is about 1%. The probability of getting 3 or more calls is:

``1-.8187-.1637-.0164=.0012

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The probability of getting more than two calls in a random 1 hour period is approximately .0012

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Using a TI-83 we can use poissoncdf(.2,2) to get .9988515188 and 1-.9988515188=.0011484812 which agrees very well.