Thread: Poisson

I need some help. I just started a class because they don't offer the prerequisite this quarter. If someone knows where I can find a good intro to Poisson process document/webpage/etc that would also be helpful.


Phone calls arrive according to a poisson process with a rate of 3 calls per minute. Each call lasts exactly 10 seconds. Find the probabilities
1) that a call does not overlap with either the call before or after itself
2) that a call does not overlap with any other call

L=.5 calls/10sec
x=2 (a second call arriving within the 10s interval)
P(2,.5) = (e^-.5)(.5^2)/2! = 0.0758
1) 2*P(2,.5) ----once for the chance of starting while the previous call is occurring, once for the chance of a call starting while your call is going
Is this even close?

2) ??

Originally Posted by chris12

I need some help. I just started a class because they don't offer the prerequisite this quarter. If someone knows where I can find a good intro to Poisson process document/webpage/etc that would also be helpful.


Phone calls arrive according to a poisson process with a rate of 3 calls per minute. Each call lasts exactly 10 seconds. Find the probabilities
1) that a call does not overlap with either the call before or after itself
2) that a call does not overlap with any other call

L=.5 calls/10sec
x=2 (a second call arriving within the 10s interval)
P(2,.5) = (e^-.5)(.5^2)/2! = 0.0758
1) 2*P(2,.5) ----once for the chance of starting while the previous call is occurring, once for the chance of a call starting while your call is going
Is this even close?

2) ??

For (1) I would use the fact that the intervals between arivals for a Poisson process have an exponential distribution, and the interval lengths are independent.

Then you just have to find the probability that the previous interval was >10s, and that the following interval also was >10s. As these events are independednt we then just multiply these probabilities together.

(2) has me puzzled, it seems indistinguishable from (1).

RonL

So...
L=3
x=1/6

F(x;L)=1-e^-Lx
1-e^-3*1/6
1-e^-.5

For an overlap of the next call.

a) (1-e^-.5) + (1-e^-.5) because each event is independent and I want the chance of either happening
b) same probability because to overlap with any call you must first overlap with either the previous or next call

Originally Posted by chris12

So...
L=3
x=1/6

F(x;L)=1-e^-Lx
1-e^-3*1/6
1-e^-.5

For an overlap of the next call.

a) (1-e^-.5) + (1-e^-.5) because each event is independent and I want the chance of either happening
b) same probability because to overlap with any call you must first overlap with either the previous or next call

That looks about right.

RonL