Answer to Question #99865 in Statistics and Probability for Linta

Question

a company uses 3 machines A,B,C.propotion of products produced by the machine are .2,.48,.22 . the percent defective for each machines is 1% 2% and 3% of their individual output. a defective output is drawn from a days production.what is probability that it was not prpduced by A?.

Answer 1

Suppose E is the event of getting defected output.

Then from the question, we have:

$P(A)=0.2,P(B)=0.48,P(C)=0.22$

$P(E/A)=0.01,P(E/B)=0.02,P(E/C)=0.03$

Now, we need to find the probability that if the bulb is defective then it is not produced by A, that means it is produced by either B or C.

That is, we need to find $P(B/E)+P(C/E)$

Now,

$P(B/E)=(P(B)*P(E/B))/(P(A)*P(E/A)+P(B)*P(E/B)+P(C)*P(E/C))$

$P(B/E)=(0.48*0.02)/(0.2*0.01+0.48*0.02+0.22*0.03)$

$=0.0096/(0.0020+0.0096+0.0066)$

$=0.0096/0.0182$

$=0.5275$

Now, similarly:

$P(C/E)=(P(C)*P(E/C))/(P(A)*P(E/A)+P(B)*P(E/B)+P(C)*P(E/C))$

$P(C/E)=(0.22*0.03)/(0.2*0.01+0.48*0.02+0.22*0.03)$

$=0.0066/(0.0020+0.0096+0.0066)$

$=0.0066/0.0182$

$=0.3626$

So, the probability that the if the output is defected then it was not produced by A will be equal to ,

$P(B/E)+P(C/E)=0.5275+0.3626$

$=0.8901$