Question

A reservation service employs five information operators who receive requests for information independently of one another, each according to a Poisson process with rate μ = 2 per minute. A. What is the 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, the first operator receives 2 or fewer requests?

Answer 1

Answer: 1) 0.13534, or 13.534% 2) 0.67668, or 67.668%

Explanation:

QUESTION #1:

The probability that the first operator receives no requests during a given 1-minute period can be calculated using the Poisson distribution formula. In this case, the rate parameter, λ, is given as 2 requests per minute for each operator.

The formula for the Poisson distribution is P(x; λ) = (e^(-λ) * λ^x) / x!, where P(x; λ) is the probability of x events occurring in a given interval, λ is the average rate of events, e is the base of the natural logarithm (approximately 2.71828), and x! denotes the factorial of x.

To find the probability that the first operator receives no requests (x=0), we substitute x=0 and λ=2 into the formula: P(0; 2) = (e^(-2) * 2^0) / 0!

Since 0! equals 1, the denominator becomes 1: P(0; 2) = e^(-2) * 1 e^(-2) is approximately 0.13534.

Therefore, the probability that the first operator receives no requests during a given 1-minute period is approximately 0.13534, or 13.534%.

QUESTION #2:

To find the probability that the first operator receives 2 or fewer requests during a given 1-minute period, we can sum up the individual probabilities for each possible number of requests: 0, 1, and 2.

The rate parameter, λ, is given as 2 requests per minute for each operator, so we will use the Poisson distribution formula to calculate the probabilities.

The formula for the Poisson distribution is P(x; λ) = (e^(-λ) * λ^x) / x!, where P(x; λ) is the probability of x events occurring in a given interval, λ is the average rate of events, e is the base of the natural logarithm (approximately 2.71828), and x! denotes the factorial of x.

To find the probability that the first operator receives 0, 1, or 2 requests, we calculate the individual probabilities and sum them up:

P(0; 2) = (e^(-2) * 2^0) / 0! = e^(-2) * 1 P(1; 2) = (e^(-2) * 2^1) / 1! = 2e^(-2) P(2; 2) = (e^(-2) * 2^2) / 2! = 2e^(-2) Adding up these probabilities: P(0; 2) + P(1; 2) + P(2; 2) = e^(-2) + 2e^(-2) + 2e^(-2) = 5e^(-2)

Approximately, 5e^(-2) is 0.67668.

Therefore, the probability that the first operator receives 2 or fewer requests during a given 1-minute period is approximately 0.67668, or 67.668%.