Question

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.)

Answer 1

Answer:

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson process with rate 2 per minute

This means that $\mu = 2$

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).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.135$

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 $n = 6$

Each operator has a 13.5% probability of receiving no requests during a minute, so $p = 0.135$

This is P(X = 3).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{6,3}.(0.135)^{3}.(0.865)^{3} = 0.03185$

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