State Bayes’ theorem. In a bolt factory, three machines A,B and C
manufacture 25% , 35% and 40% of the total product respectively. Of
these outputs 5% , 4% and 2% respectively, are defective bolts. A bolt
is picked up at random and found to be defective. What are the
probabilities that it was manufactured by machines A,B and C?
Answers
Answer
Answer
Consider the problem
Let
A: bolt manufactured from machine A
B: bolt manufactured from machine B
C: bolt manufactured from machine C
D: bolt is defective
WE need find the probability that the bolt is manufactured by machine AorC and machine B, if it is defective.
That is P( B/D )
So, P(B/D )=
P(A).P( D/A )+P(B).P(D/B )+P(C).P(D/C )
P(A)= probability that the bolt is made by machine A
=25%=
100
25
=0.25
P(B)= probability that the bolt is made by machine B
=35%=
100
35
=0.35
P(C)= probability that the bolt is made by machine C
=40%=
100
40
=0.40
P(AorC)= probability that the bolt is made by machine AorC
=65%=
100
65
=0.65
And,
P(D/A )= probability of a defective bolt from machine A
=5%=
100
5
=0.05
P(D/B )= probability of a defective bolt from machine B
=4%=
100
4
=0.04
P( D/C)= probability of a defective bolt from machine C
=2%=
100
2
=0.02
P(D/A or C )= probability of a defective bolt from machine A or C
=0.07
Now apply Bayes' Theorem
P(B/D)= 0.35×0.04 / 0.25×0.05+0.35×0.04+0.04×0.02 =
0.014 / 0.0125+0.014+0.008 = 0.014 / 0.0345 =
140 / 345 = 28 / 69
Therefore, required probability by the machine B is
28 / 69
And Machine A or C is
Apply Bayes' Theorem
P(A or C / D )= 0.65×0.07 / 0.65×0.07+0.35×0.04
= 13 / 17
Answer:
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