Answer to Question #100094 in Statistics and Probability for Tanveer

Question

A garment factory produces certain types of output by three machines.The respective daily productions were : Machine A = 4500 units, Machine B =2500 units, Machine C = 3000 units. According to past experience it is found that the fractions of defective output produced by the three machines were 1%,1.2% and 2% RESPECTIVELY.An Item is drawn at random free the day's production run and is found to be defective. What is the probability that it comes from the output of i) Machine A;(ii) Machine B;(iii) Machine C

Answer 1

Let $D$ be the event '' defective item is chosen,'' $A$ be the event '' item is produced by machine A'', $B$ be the event '' item is produced by machine B'', and $C$ be the event '' item is produced by machine C''.

$$P(A)={4500 \over 4500+2500+3000}=0.45$$

$$P(B)={2500 \over 4500+2500+3000}=0.25$$

$$P(C)={3000 \over 4500+2500+3000}=0.3$$

Conditional probabilities

$$P(D|A)=0.01, P(D|B)=0.012, P(D|C)=0.02$$

i) From Bayes theorem probability of defective item from machine A

$$P(A|D)={P(A)P(D|A) \over P(A)P(D|A)+P(B)P(D|AB)+P(C)P(D|C)}=$$

$$={0.45(0.01) \over 0.45(0.01)+0.25(0.012)+0.3(0.02)}\approx0.0296$$

ii) Probability of defective item from machine B

$$P(B|D)={P(B)P(D|B) \over P(A)P(D|A)+P(B)P(D|AB)+P(C)P(D|C)}=$$

$$={0.25(0.012) \over 0.45(0.01)+0.25(0.012)+0.3(0.02)}\approx0.0222$$

iii) Probability of defective item from machine C

$$P(C|D)={P(C)P(D|C) \over P(A)P(D|A)+P(B)P(D|AB)+P(C)P(D|C)}=$$

$$={0.3(0.02) \over 0.45(0.01)+0.25(0.012)+0.3(0.02)}\approx0.0444$$