# What is the probability the phone will ring while the desk employee is gone on his break? Phone calls arrive at the rate of 15 per hour at the reservation desk for a hotel, according to a poisson distribution.If no calls are currently being processed, and the desk employee leaves the desk for a 10-minute break,

The phone calls arrive at the rate of 15 per hour. There is one phone call arriving on an average every four minutes. 10 minutes is 2.5 times this interval.

The probability of an event occurring k number of times in a given interval lambda is `P(k, lambda) = (lambda^k*e^(-lambda))/(k!)`....

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

The phone calls arrive at the rate of 15 per hour. There is one phone call arriving on an average every four minutes. 10 minutes is 2.5 times this interval.

The probability of an event occurring k number of times in a given interval lambda is `P(k, lambda) = (lambda^k*e^(-lambda))/(k!)`. The probability of the phone not ringing while the employee is on his 10 minute break is `P(0, lambda)` . The required probability that the phone will ring is `1 - P(0, lambda)`.

Determining  `P(k, lambda) = (lambda^k*e^(-lambda))/(k!)` with `lambda = (15/60)*10 = 2.5` and k = 0 gives:

`(lambda^k*e^(-lambda))/(k!)`

=> `(2.5^0*e^-2.5)/1`

=> `e^-2.5`

=> 0.08208

The probability that the phone rings while he is gone is 1 - 0.08208 = 0.918

The required probability is  0.918