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
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P(B1) P(A | B1)
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(i) P(B1 | A) = |
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P(B1) P(A | B1)
+ P(B2) P(A | B2) + P(B3)
P(A | B3)
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(25/100) X (5/100)
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or P(B1 | A) = |
|
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[(25/100) X (5/100)] + [(35/100)
X (4/100)] + [(40/100) X (2/100)]
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or P(B1 | A) = 25/69
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P(B3) P(A | B3)
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(ii) P(B3 | A) = |
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P(B1) P(A | B1)
+ P(B2) P(A | B2) + P(B3)
P(A | B3)
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(40/100) X (2/100)
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or P(B1 | A) = |
|
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[(25/100) X (5/100)] + [(35/100)
X (4/100)] + [(40/100) X (2/100)]
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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
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P(X Ç
S)
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(i) P(X | S) = |
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P(S)
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(90/10000)
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or P(X | S) = |
|
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(335/10000)
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or P(X | S) = 18/67
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P(Y Ç
S)
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(ii) P(Y | S) = |
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P(S)
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(140/10000)
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or P(Y | S) = |
|
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(335/10000)
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or P(Y | S) = 28/67
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