Question

A computer undergoes downtime if a certain critical component fails. This component is known to fail at an average rate of once per five weeks according to the Poisson distribution. No significant downtime occurs if replacement components are on hand because repair can be made rapidly. There are two components on hand, and ordered replacements are not due for eight weeks.

What is the probability of significant downtime occurring before the ordered components arrive?

If the shipment is delayed two weeks, what is the probability of significant downtime occurring before the shipment arrives?

Answer 1

Average failure rate = 1/5 per week = 8/5 per 8 weeks = 1.6 per 8 weeks

Let X be the number of failures in 8 weeks.

X ~ Poisson( = 1.6)

probability of significant downtime = P(X > 2)

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

= 1 - 0.2019 - 0.3230 - 0.2584

= 0.2167

Average failure rate = 1/5 per week = 10/5 per 10 weeks = 2 per 10 weeks

Let Y be the number of failures in 10 weeks.

Y ~ Poisson( = 2)

If the shipment is delayed two weeks, probability of significant downtime = P(Y > 2)

= 1 - (P(Y = 0) + P(Y =1) + P(Y =2))

= 1 - 0.1353 -  0.2707 - 0.2707

= 0.3233