Question

Certain electrical disturbances occur according to a Poisson process with rate 3 per hour. These disturbances cause damage to a computer.

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?

c) Assume that a crash will not happen unless there are two disturbances within 5 minutes of each other. Calculate the probability that the computer will crash in the coming 10 minutes.

Answer 1

    The disturbance rate is 3 per hour, that is 0.5 per 10 minutes                      
   Let X be the number of disturbances in 10 minutes                      
   X follows a Poisson distribution with λ = 0.5 disturbances per 10 minutes                      
   The pdf of the Poisson distribution is                       
   That is                       
                         
                          
a)    A single disturbance can crash the computer                      
   To find P(system will crash in 10 minutes)                      
   that is to find P(there is atleast 1 disturbance in 10 minutes)                      
   that is to find P(X ≥ 1)                      
   P(X ≥ 1) = 1- P(X < 1)                      
                    = 1 - P(X = 0)                      
   We use the Excel function POISSON.DIST to find the probability                      
   P(X ≥ 1) = 1 - POISSON.DIST(0, 0.5, FALSE)                      
                    = 1 - 0.6065                      
                    = 0.3935                      
   P(system will crash in 10 minutes) = 0.3935                      
                          
b)    Two disturbances can crash the computer                      
   To find P(system will crash in 10 minutes)                      
   that is to find P(there are 2 or more disturbances in 10 minutes)                      
   that is to find P(X ≥ 2)                      
   P(X ≥ 2) = 1- P(X < 2)                      
                    = 1 - [P(X = 0) + P(X = 1)]                      
   We use the Excel function POISSON.DIST to find the probability                      
   P(X ≥ 1) = 1 - [POISSON.DIST(0, 0.5, FALSE) + POISSON.DIST(1, 0.5, FALSE)]                      
                    = 1 - 0.9098                      
                    = 0.0902                      
   P(computer will crash in 10 minutes) = 0.0902                      
                          
c)    For a Poisson process, the time between 2 events follows an exponential distribution                      
   Let Y be the time between the 2 disturbances                      
   For the Poisson process mean = 0.5 disturbances per 10 minutes                      
   which is 0.05 disturbances per minute                      
   Y follows Exponential distribution with λ = 0.05                      
   Thus, we have to find P(Y < 5), where Y is the time between 2 disturbances                      
   We use Excel function EXPON.DIST to find the probability                      
   P(Y < 5) = EXPON.DIST(5, 0.05, TRUE)                      
                   = 0.2212                      
   P(computer will crash in 10 minutes) = 0.2212