A reservation service employs six information operators who receive requests for information independently of one another, each according to a Poisson process with rate = 2 per minute. a. What is the probability that during a given 1 min period, the first operator receives no requests? (Round your answer to three decimal places.) b. What is the probability that during a given 1 min period, exactly three of the six operators receive no requests? (Round your answer to five decimal places.)

a) 0.135 = 13.5% probability that during a given 1 min period, the first operator receives no requests.

b) 0.03185 = 3.185% probability that during a given 1 min period, exactly three of the six operators receive no requests

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

In which

x is the number of sucesses

e = 2.71828 is the Euler number

is the mean in the given time interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

Poisson process with rate 2 per minute

This means that

a. What is the probability that during a given 1 min period, the first operator receives no requests?

Single operator, so we use the Poisson distribution.

This is P(X = 0).

0.135 = 13.5% probability that during a given 1 min period, the first operator receives no requests.

b. What is the probability that during a given 1 min period, exactly three of the six operators receive no requests?

6 operators, so we use the binomial distribution with

Each operator has a 13.5% probability of receiving no requests during a minute, so

This is P(X = 3).

0.03185 = 3.185% probability that during a given 1 min period, exactly three of the six operators receive no requests