Wrecking my brains on this Poisson Distribution question.

[aznking1]

Hi. I normally can solve poisson distribution questions with ease. But this one question had me thinking for hours on end with no solution. It would be great if someone can help me.

QN:

The number of incoming calls per minute, X, to a telephone exchange has a Poisson distribution with mean 2. The number of outgoing calls per minute, Y, has an independent Poisson Distribution with mean 1.

In a randomly chosen 1-minute period, find the probability that there are more incoming calls than outgoing calls, given that there is a total of AT LEAST 3 calls.

My Working:

Let incoming calls > outgoing calls be A
Let a total of at least 3 calls be B

P(A|B) = P(A∩B)/P(B)

The part where i am stuck is finding P(A∩B), i.e, Probability of more incoming calls than outgoing calls AND a total of at least 3 calls. There are so many cases. Could be 3 incoming calls, 0 outgoing calls. Could be 4 incoming calls 1 outgoing calls. E.t.c. Then you take the sum of all these cases and it would be = P(A∩B).

But the number of cases are endless. So, just how am i supposed to calculate P(A∩B)? I am at my wits ends!

The answer is 0.705 if you manage to solve it.

 

[Jameson]

Hi aznking2,

Welcome to MHB! :)

Since this deals with Poisson random variables, there is no upper limit on the frequency of events so we can't calculate it that way. My guess is that we'll have to use the property the $P(A)=1-P(A')$ because we can calculate the opposite what of what we want more easily.

Can you please verify that .705 should be the answer?

 

[aznking1]

Hi! Thanks for the quick reply.

I have thought of using P(A)=1−P(A′). However, it is not as straightforward as it seems. What exactly is your A' over here? A would be the case of more incoming calls than outgoing calls AND a total of of at least 3 calls. What would be A'? The phrasing eludes me and i have been trying to figure it out. If I figure it out I believe I would be able to solve it.

I do not think A' would be the case of more incoming calls than outgoing calls AND a total of less than 3 calls? So confusing.

And yes, I checked the answer of the book and it is indeed 0.705.