Exponential Distribution and probability

[6021023]

Homework Statement

The time between calls to a corporate office is exponentially distributed with a mean of 10 minutes. What is the probability that there are more than three calls in one-half hour?

Homework Equations

F(x) = P(X <= x) = 1 - e^-(lamba*x)

The Attempt at a Solution

P(X > 3) = 1 - P(X < 3)
= 1 - [1 - e^-(10*3)]
=e^-(10*3)

I think this answer is wrong though because I think that P(X > 3) actually means the probability that at least 30 minutes will pass before the first call instead of the probability that there are more than three calls in one half hour, but I'm not sure.

 

[John Creighto]

Homework Statement

The time between calls to a corporate office is exponentially distributed with a mean of 10 minutes. What is the probability that there are more than three calls in one-half hour?

Homework Equations

F(x) = P(X <= x) = 1 - e^-(lamba*x)

P(X <= x) is not the notation for a probability distribution. It is the notation for a cumulative distribution. Additionaly I think the PDF is probably just:

e^-(lamba*x)

I think this answer is wrong though because I think that P(X > 3) actually means the probability that at least 30 minutes will pass before the first call instead of the probability that there are more than three calls in one half hour, but I'm not sure.

The question gave you the distribution for the average time between calls. However, you want the amount of calls that occurred in a 10 minute interval. It sounds like a bit of a conditional probability question.

In the 10 minute intervals you can have three calls anywhere in the interval. This will be a three dimensional space of random variables a b c, where a,b,c represent when the call occurred in the interval.

You need:
P(a|b|c)

I know:

P(a|b)=P(a&b)/P(b)

You need to find a more general expression for this. Then you need to integrate P(a|b|c) where a,b,c Go from zero minutes to 10 minutes.

 

[6021023]

Looking at a solution to a similar problem in the book shows them solving it by treating it as a Poisson random variable.

 

[Mark44]

P(X > 3) = 1 - P(0 <= X <= 3) = 1 - P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Now, all you need is calculate the four probabilities on the right, add them, and subtract the total from 1.

Mark