Question

Consider a computer system with Poisson job-arrival stream at an average of 10 jobs per 4 minutes. Determine the probability that in any one-minute interval there will be (20 points):

1. 1. Exactly 5 jobs
2. 2. At least 3 jobs
3. 3. Between 2 to 5 jobs inclusive

Answer 1

Job arrival = 10 jobs per 4 minutes

Lambda = Job arrival rate per minute = 10 / 4 = 2.50

Lambda = 2.50

Poisson Distribution: Probability (X = x) = e^-Lambda * Lambda^x / x! | where X is number of events to occur in a fixed-time interval.

where Lambda is the average job arrival rate and x = 0,1,2... infinity

a) Exactly 5 jobs

For exactly 5 jobs, X = 5

P (X = 5) = e^-2.5 * (2.5⁵) / 5! = 0.0821 * 97.6563 * 1 / 120

P (X = 5) = 0.0668 or 6.68%

Hence, Probability of Exactly 5 jobs in any one-minute interval is 6.68% or 0.0668.

b) At least 3 jobs

For at least 3 jobs, X >= 3 | X should be greater than equal to 3. We can find the Probability by first finding the cumulative probability of 2 jobs (which includes 1 job's probability) and subtract it from 1 to find probability of 3 or more.

P (X >= 3) = 1 - (P (X = 2) + P (X = 1) + P (X = 0))

P (X = 2) = e^-2.5 * (2.5²) / 2! = 0.0821 * 6.25 / 2

P (X = 2) = 0.2565

P (X = 1) = e^-2.5 * (2.5¹) / 1! = 0.0821 * 2.5 / 1

P (X = 1) = 0.2052

P (X = 0) = e^-2.5 * (2.5⁰) / 0! = 0.0821

P (X = 0) = 0.0821

Now putting values of P (X = 2) and P (X = 1) in P (X > = 3) formula.

P (X >= 3) = 1 - (0.2565 + 0.2052 + 0.0821) = 1 - 0.5438

P (X >= 3) = 0.4562 or 45.62%

Hence, Probability of at least 3 jobs in any one-minute interval is 45.62% or 0.4562.

c) Between 2 and 5 jobs inclusive

For jobs between 2 and 5, We need to calculate P (2 >= X <=5)

As We need at least 5, we can calculate Cumulative Probability of 5 and then exclude Probability of 1 and 0 to keep only from 2 to 5.

P (2 >= X <= 5) = Probability of at least 5 - Probability of 1 - Probability of 0

P (2 >= X <= 5) = P (X = 5) + P (X = 4) + P (X = 3) + P (X = 2) + P (X = 1) + P (X = 0) - P (X = 1) - P (X = 0)

P (2 >= X <= 5) = P (X = 5) + P (X = 4) + P (X = 3) + P (X = 2)

P (X = 5) = e^-2.5 * (2.5⁵) / 5! = 0.0821 * 97.6563 * 1 / 120 = 0.0668

P (X = 4) = e^-2.5 * (2.5⁴) / 4! = 0.0821 * 39.0625 * 1 / 24 = 0.1336

P (X = 3) = e^-2.5 * (2.5³) / 3! = 0.0821 * 15.6250 * 1 / 6 = 0.2138

P (X = 2) = e^-2.5 * (2.5²) / 2! = 0.0821 * 6.25 * 1 / 2 = 0.2565

P (2 >= X <= 5) = 0.0668 + 0.1336 + 0.2138 + 0.2565

P (2 >= X <= 5) = 0.6707 or 67.07%

Hence, Probability of between 2 and 5 jobs inclusive in any one-minute interval is 67.07% or 0.6707.