An engineer gathered data on failures of a circuit that has been considered for
ID: 3208069 • Letter: A
Question
An engineer gathered data on failures of a circuit that has been considered for use in a satellite. Alter several experiments, the engineer decided that failures could be divided into the following mutually exclusive and exhaustive categories: A = integrated circuit failure. B = PC board failure. C = power supply failure; and D = other failure. The engineer also isolated a common problem M that is thought to contribute to all four types of failure. He established the following connections; P(A) - 0.40;P(M l A) - 0.05; P(B) = 0.25; P(M l B) = 0.05; P(C) = 0.30; P(A | C) = 0.04; P(D) = 0.05;P(A | D) = 0.01. Find P(M). P(M'), P(C M). P(C M'). Interpret these events within then engineering context. The probability of producing a defective item is 0.02. What is the probability that zero out of 10 items that are selected independently are defective? How many defective items do you expect in a sample of 100 items?Explanation / Answer
probability of common problem =P(M)=P(A)*P(M|A)+P(B)*P(M|B)+P(C)*P(M|C)+P(D)*P(M|D)
=0.4*0.05+0.25*0.05+0.30*0.04+0.05*0.01=0.045
hence probabilty of not a common problem =P(M')=1-0.045=0.955
probability of power supply faiure given common problem happened =P(C|M)=P(C)*P(M|C)/P(M)=0.3*0.04/0.0455
=0.2637
probability of power supply faiure given common problem not happened =P(C|M')=P(C)*P(M'|C)/P(M')
=0.3*0.96/0.955=0.3015
4)
a)probabilty of 0 defect out of 10 =P(no defect in all 10) =(1-0.02)10=0.817
b)expected defects=np=100*0.02=2