Failures occur for a mechanical process according to a Poisson process. Failures

Question

Failures occur for a mechanical process according to a Poisson process. Failures are classified as cither major or minor. Major failures occur at a rate of 1.5 failures per hour. Minor failures occur at a rate of 3.0 failures per hour. (a) Find the probability that two failures occur in 1 hour (b) Find the probability that in half an hour, no major failure occurs. (c)Find the probability that in 2 hours, at least two major failures occur or at least two minor failures occur (d) For major failure, find the probability that the time interval between two failures is between half an hour and one hour.

Explanation / Answer

Solution:

(a)

Major failures at thebrate of 1.5 failures per hour.
Minor failures at thebrate of 3 failures per hour.
Probability that 2 failures coours in 1 hour = e-x /x!

= (1.5)1*e-1.5 /1! + (1.5)2*e-1.5 /2! + (3)1*e-3 /1! + (3)2*e-3 /2! = 0.5249

(b)

Probability that in half an hour, no major failures occurs = e-x /x!

Here = 1.5, x = 0

Thus, Required probability = (1.5)0*e-1.5 /0! = e-1.5 = 0.223

(c)

For two hours expected numebr of major failures =1.5 * 2 =3

and expected numebr of minor failures = 2 * 3 =6

Probability that in 2 hours, at least two major failures occur = P(X1)

                                                                                              = 1 - P(at most 1 major failure)

                                                                                              = 1 - e-x /x! = 0.8008

Again, the probability that in 2 hours, at least two minor failures occur = P(X2)

                                                                                                               =1-e-x /x! = 0.9826

Hence  probability that in 2 hours, at least two major failures occur or at least two minor failures occur

= P(X1 U X2) = P(X1) + P(X2) - P(X1)*P(X2) =0.9965

This can also be solved by probability =1-P( none of major or minor events happened more then once)

(d)

From exponential distribution ; average time for failures =1/1.5 hour

From above probability that the time interval between two failures is between half an hour and one hour

= P(0.5 <X<1) = e-0.5*1.5 -e-1*1.5 = 0.2492

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