Q. 12. In a bolt factory machines A, B and C manufacture respectively 25, 30 and 40 percent of the total. Out of their total outputs 5, 4 and 2 percent are defective. A bolt is drawn from the produce at random and is found to be defective. What is the probability that it is manufactured by (i) factory A (ii) factory C? (June 99)

Solution. This problem is based on Bayes theorem.
Let A = an event that the output is defective
B1 = an event that bolt is manufactured by machine A
B2 = an event that bolt is manufactured by machine B
B3 = an event that bolt is manufactured by machine C

Therefore, P(B1) = 25/100, P(B2) = 35/100, P(B3) = 40/100
P(A | B1) = 5/100, P(A | B2) = 4/100, P(A | B3) = 2/100

 
P(B1) P(A | B1)
(i) P(B1 | A) =
 
P(B1) P(A | B1) + P(B2) P(A | B2) + P(B3) P(A | B3)

 
(25/100) X (5/100)
or P(B1 | A) =
 
[(25/100) X (5/100)] + [(35/100) X (4/100)] + [(40/100) X (2/100)]

or P(B1 | A) = 25/69

 
P(B3) P(A | B3)
(ii) P(B3 | A) =
 
P(B1) P(A | B1) + P(B2) P(A | B2) + P(B3) P(A | B3)

 
(40/100) X (2/100)
or P(B1 | A) =
 
[(25/100) X (5/100)] + [(35/100) X (4/100)] + [(40/100) X (2/100)]

or P(B3 | A) = 16/69

Q. 13. In a bolt factory machines A, B and C manufacture respectively 30, 35 and 35 percent of the total. Out of their total outputs 3, 4 and 3 percent are defective. A bolt is drawn at random and is found to be defective. What is the probability that it is manufactured by (i) factory A (ii) factory B? (June 2002)

Solution.
Let X, Y and Z denote the event that a bolt is manufactured by the machine A, B and C respectively.
Let S denotes the event that the bolt is drawn now.

P(X Ç S) = (30/100) X (3/100) = 90/10000
P(Y Ç S) = (35/100) X (4/100) = 140/10000
P(Z Ç S) = (35/100) X (3/100) = 105/10000

P(S) = [(90/10000) + (140/10000) + (105/10000)]
or P(S) = 335/10000

 
P(X Ç S)
(i) P(X | S) =
 
P(S)

 
(90/10000)
or P(X | S) =
 
(335/10000)

or P(X | S) = 18/67

 
P(Y Ç S)
(ii) P(Y | S) =
 
P(S)

 
(140/10000)
or P(Y | S) =
 
(335/10000)

or P(Y | S) = 28/67



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