# How many clients do the bankers meet? (Poisson Process)

A manager looks at how many new clients his bankers meet.

He assumes that for every 5 consecutive minutes in a meeting, the banker each meets a new client. He chooses to look at three of his bankers, $A,B$ and $C$. The following data is provided in time.

Banker $A$: $09:00 - 09:07, 09:12 - 09:38, 12:40-12:45$

Banker $B$: $08:35-09:45, 16:35-17:00$

Banker $C$: $10:00-12:00$

Then assuming that clients are met with some average rate, say, $\lambda$, we can model this as a Poisson process for each banker.

Estimating the parameter $\lambda$ with how many clients the banker met divided by the total time elapsed, we have:

$A$ ~ $Pois(\frac{7}{38}t)$ in $38$ minutes per unit of time.

$B$ ~ $Pois(\frac{1}{5}t)$ in $95$ minutes per unit of time.

$C$ ~ $Pois(\frac{1}{5}t)$ in $120$ minutes per unit of time.

We can standardise the date to get one hour per unit of time.

Then we have:

$A$ ~ $Pois(0.290858*t)$

$B$ ~ $Pois(0.126315*t)$

$C$ ~ $Pois(0.1*t)$

Calculate the probability that all 3 bankers meet more than 2 clients in 3 hours, individually and then as a group (together).

We look at individiually first.

We have $\mathbb{P}(A(3)>2) + \mathbb{P}(B(3)>2) + \mathbb{P}(C(3)>2)$

$= 3-\exp(-3*0.290858) \sum_{k=0}^{2} \frac{(3*0.290858)^k}{k!} -\exp(-3*0.126315) \sum_{k=0}^{2} \frac{(3*0.126315)^k}{k!} -\exp(-3*0.1) \sum_{k=0}^{2} \frac{(3*0.1)^k}{k!} \approx 0.06856$.

As a group it is (seemingly) a nightmare to solve as we have to look at all kinds of combinations.

We have

$\mathbb{P}(A(3) \geq 1) * \mathbb{P}(B(3) \geq 1) * \mathbb{P}(C(3) \geq 1) + \mathbb{P}(A(3)> 2) * \mathbb{P}(B(3)=0) * \mathbb{P}(C(3)=0)+ \mathbb{P}(A(3)=0)\mathbb{P}(B(3)>2)\mathbb{P}(C(3)=0) + \cdots$

i.e. All possible combinations. Is there a shorter way to get the answer here? I have a lot of probability expressions to evaluate.

Now suppose that 3 types of clients can be met. Namely, managers, analysts and interns. Managers are met with probability $\frac{3}{5}$, analysts are met with probability $\frac{1}{2}$ and interns with probability $\frac{1}{5}$. Suppose that in any meeting the bankers are in, there are $50$ people, all of the type as explained above, in the meeting.

Calculate the probability that over $5$ hours, bankers $A$ and $B$ each meet only $3$ analysts and $3$ interns.

We want $\frac{3}{15}(\mathbb{P}(A(5)=3) + \mathbb{P}(B(5)=3)) + \frac{3}{25}(\mathbb{P}(A(5)=3)+\mathbb{P}(B(5)=3)) \approx 0.04546$.

I would appreciate some verification on my answers; are they correct? Is there a shorter way to obtain the answer?

A lot of understanding how to model a real life process is down to understanding the data and using sensible assumptions.

-