This is a pretty hard problem. I have answer to a simpler problem. I will think about a full answer later.

The simpler problem is: the Calls to the tech support center arrive in a Poisson process with a rate of 20 per hour. What is the probability that there will be more than 3 Calls to the support center in any minute?

Poisson distribution with a mean of 20/hr, gives mean=0.3/min. So, to have more than 3 incoming calls within a given minute is:

P(x>=4) = 1-P(0)-P(1)-P(2)-P(3) = 0.00002.

The Poisson Distribution is: P=mean**(i) e**(-mean)/i! , i =0, 1, 2, 3, 4, ...

i=0, p[0]=0.74081837017575, #probability of no calls coming in a min
i=1, p[1]=0.222245511052725, #probability of 1 calls coming in a min
i=2, p[2]=0.0333368266579087, #probability of 2 calls coming in a min
i=3, p[3]=0.00333368266579087, #probability of 3 calls coming in a min
i=4, p[4]=0.000250026199934316, #prob. of 4 calls coming in a min.
i=5, p[5]=1.50015719960589e-05, #prob. of 5 calls coming in a min.
i=6, 7, 8......... odds are very small. all adds up is <0.0000001.

NOW THE UNANSWERED PART IS HOW THE EXPONENTIAL DISTRIBUTION WITH MEAN 3MIN PLAY INTO THIS PROBLEM.