# How likely is it that exactly 1 customer will arrive in any 5 minutes? If there are no additional customers waiting (or walking up to the counter) when a particular customer begins her transaction, what is the probability that no one will be waiting in line when her transaction is finished?

People arrive a a particular sales counter at the rate of 6 in any 15 minute period and are served on a "first come first served" basis. Arrivals occur randomly throughout the hour and the arrival rate is the same during the entire business day. It takes 5 minutes to service a single customer. How likely is it that exactly 1 customer will arrive in any 5 minutes?

Using the basic informaion from above: If there are no additional customers waiting (or walking up to the counter) when a particular customer begins her transaction, what is the probability that no one will be waiting in line when her transaction is finished?

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...l P (1) in any 5 minutes using Poisson distribution

Poisson distribution:

P(x)= (lambda ^ x) (e ^-lambda ) / x!

where

P(x)= probability of x

lambda = average

^ is raised to the power of , e is the exponential function , x! is x factorial

We have 6 arrivals in 15 minutes

Therefore, in a 5 minute interval there would be an average of 2 arrivals =(5 / 15) x 6 ...