Question

Assume that power surges occur as a Poisson process with rate 3 per hour. These events cause damage to a computer, thus a special protecting unit has been designed. That unit now has to be removed from the system for 10 minutes for service.

(a) Assume that a single disturbance will cause the computer to crash. What is the probability that the system will crash in the coming 10 minutes?
(b) Assume that the computer will survive a single disturbance, but the second such disturbance will cause it to crash. What is, now, the probability that the computer will crash in the coming 10 minutes?

Answer 1

a) 0.30327

b) 0.07582

Explanation:

This is a Poisson distribution problem

Poisson distribution probability function is given as

P(X = x) = (e^-λ)(λˣ)/x!

where λ = mean = 3 power surges per hour

x = variable whose probability is required

a) If a single disturbance in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of a single disturbance (power surge) in 10 minutes?

λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.

P(X = x) = (e^-λ)(λˣ)/x!

λ = 0.5 disturbance per 10 minutes

x = 1 disturbance per 10 minutes

P(X = 1) = (e^-0.5)(0.5¹)/1! = (e⁻⁰•⁵)(0.5)/1!

= 0.30327

b) If two disturbances in 10 minutes will crash the computer, what is the probability of the computer crashing, that is probability of two disturbances (power surges) in 10 minutes?

λ = 3 disturbances per hour = 1 disturbance per 20 minutes = 0.5 disturbance per 10 minutes.

P(X = x) = (e^-λ)(λˣ)/x!

λ = 0.5 disturbance per 10 minutes

x = 2 disturbance per 10 minutes

P(X = 2) = (e^-0.5)(0.5²)/2! = (e⁻⁰•⁵)(0.25)/2!

= 0.07582

Hope this Helps!!!