Poisson probability

**galactus** · December 14th 2009, 08:42 AM

Hi All:

This time I am asking a question. I am teaching stats next semester after several years of not doing it. I was looking through an old stats book and doing some problems. Here is a Poisson that has me stumped.

"Cars going through an intersection average 40 per hour. What is the probability that the time between the arrival of a car and the second car after it is at least 2 minutes?"

In this case, they are asking for 'at least 2 MINUTES", but the average is given in hours. So, ${\lambda}=\frac{40}{60}=\frac{2}{3}$

I was thinking maybe a gamma, like so:

$\frac{1}{(\frac{2}{3})^{2}}\int_{2}^{\infty}te^{\f rac{-t}{\frac{2}{3}}}dt=.199$

but that does not seem correct.

$\frac{2}{3}\int_{2}^{\infty}e^{\frac{-2}{3}t}dt=e^{\frac{-4}{3}}$

But this would be fine if the problem said, "what is the probability that the time between arrivals of any two cars is at least two minutes.

It's the "arrival of a car and the second after that is at least two minutes" I am unsure of.

Any ideas what I am overlooking. I feel rather obtuse on this. MrF, what is my oversight/mental block?

**Focus** · December 14th 2009, 09:12 AM

Originally Posted by galactus

I was thinking maybe a gamma, like so:

$\frac{1}{(\frac{2}{3})^{2}}\int_{2}^{\infty}te^{\f rac{-t}{\frac{2}{3}}}dt=.199$

but that does not seem correct.

$\frac{2}{3}\int_{2}^{\infty}e^{\frac{-2}{3}t}dt=e^{\frac{-4}{3}}$

But this would be fine if the problem said, "what is the probability that the time between arrivals of any two cars is at least two minutes.

It's the "arrival of a car and the second after that is at least two minutes" I am unsure of.

Any ideas what I am overlooking. I feel rather obtuse on this. MrF, what is my oversight/mental block?

Well assuming that the cars arrive independently, this forms a Poisson process with the rate given. Poisson process has a holding time that is exponential with the same rate, and the holdings are independent both of each other and the process.

Hope this helps.

**galactus** · December 14th 2009, 03:50 PM

I think I got it.

$\text{P(0 or 1 car passes by in 2 minutes)}$

$e^{-{\lambda}t}+e^{-{\lambda}t}\cdot {\lambda}t={\lambda}^{2}\int_{2}^{\infty}te^{-{\lambda}t}dt$

When ${\lambda}=\frac{2}{3}$ , then

$\frac{7}{3}e^{-\frac{4}{3}}\approx .615$

**Focus** · December 15th 2009, 06:12 AM

By a holding time I mean how long it stays constant. Say that $\mathbf{e}_\lambda$ is an exponential r.v. with parameter $\lambda$ . So the probability you are looking for is $\mathbb{P}(\mathbf{e}_\lambda \geq 2)=e^{-2\lambda}=e^{-4/3}\approx 0.263$ .

Your last answer is obviously incorrect as the mean, which is 1.5 minutes between cars, is greater than the median. 2 minutes or more should have a probability less than a half.

I was trying to support your initial line of thinking. As the cars arrive independently, we wait an exponential time between any of them. The answer is the same for any two cars.

Disclaimer: I may be wrong as I am starting to confuse myself as well

**galactus** · December 15th 2009, 06:59 AM

Thank you for your response.

The probability I got for "what is the probability that the arrival time between any two cars is at least two minutes?" is what you have $\frac{2}{3}\int_{2}^{\infty}e^{\frac{-2}{3}t}dt=.263$

But, for "what is the probability that the arrival time between a car and the second car after it is at least two minutes?", I got:

$1-P(\text{0 or 1 car arrives in 2 minutes})$

$1-\frac{2}{3}\cdot \frac{2}{3}\int_{2}^{\infty}te^{\frac{-4t}{3}}dt=.385$

Perhaps I am wrong, but it seems logical.

**awkward** · December 15th 2009, 02:02 PM

Originally Posted by galactus

Thank you for your response.

The probability I got for "what is the probability that the arrival time between any two cars is at least two minutes?" is what you have $\frac{2}{3}\int_{2}^{\infty}e^{\frac{-2}{3}t}dt=.263$

But, for "what is the probability that the arrival time between a car and the second car after it is at least two minutes?", I got:

$1-P(\text{0 or 1 car arrives in 2 minutes})$

$1-\frac{2}{3}\cdot \frac{2}{3}\int_{2}^{\infty}te^{\frac{-4t}{3}}dt=.385$

Perhaps I am wrong, but it seems logical.

The time between cars has an Exponential distribution with $\lambda = 2/3$ per minute. You want

$P(X > 2)$ where X has that distribution.

For an Exponential distribution,
$P(X > x) = e ^ {-\lambda x}$

So...

**galactus** · December 15th 2009, 02:54 PM

Yeah, that makes more sense.

$\int_{2}^{\infty}e^{\frac{-4}{3}}dt$

**awkward** · December 15th 2009, 03:23 PM

Originally Posted by galactus

Yeah, that makes more sense.

$\int_{2}^{\infty}e^{\frac{-4}{3}}dt$

No integral, just $e^{-4/3}$ .

**galactus** · December 15th 2009, 03:36 PM

Oh, the 'at least 2' threw me. I have to look at these more. I used to know these, but it's been a while. Thanks. It's coming back, but I am embarrassed. I had to prove the derivation of this some years back, believe it or not.

What messed me up was this is the same answer I got for 'the probability that the arrival time of any two cars is at least two minutes".

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