Question

4.5. There are two common types of failure to a critical electronic element of some machinery: either component A or component B may fail. If either component fails, the machinery goes down. Component A fails according to a Poisson process with mean rate 1.1 failures per shift. (The company operates 24/7 using eight-hour shifts.) Component B fails according to a Poisson process with a mean rate of 1.2 failures per day (a) What is the probability that there will be exactly five failures of the machine within a given day? (b) What is the probability that there will be no more than one failure of the machine during the next shift? (c) It is now noon and the most recent failure occurred four hours earlier. What the is the probability that the next machine failure will occur before 6PM?

Answer 1

here number of failures follow Poisson distribution with parameter 1.1+(1.2)/3=1.5 failures/shift

a) expected failure in a day(3 shifts)=3*1.5=4.5

therefore from Poisson distribution P(X=5)=e^-4.5*4.5⁵/5! =0.1708

b)

expected number of failure in 1 shift =1.5

hence P(no more then 1 failure)=P(X<=1)=P(X=0)+P(X=1)=e^-1.5*1.5⁰/0!+e^-1.5*1.5¹/1!=0.5578

c)

in 6 Hour expected number of failure =1.5*6/8=1.125

hence P(next machine failure will occur before 6 pm)=1-P(no failure in 6 hour)

=1-e^-1.125*1.125⁰/0! =1-0.3247 =0.6753