433 illustration 1 two sets of candidates are

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4.3.3. ILLUSTRATION 1 Two sets of candidates are competing for the positions of the Board of Directors of a company. The probability that the first and second sets will win are 0.65 and 0.35 respectively. If the first set wins, the probability of introducing a new product is 0.8, and the corresponding probability if the second set wins is 0.3. What is the probability that the product will be introduced? Solution Let, A 1 = the events of first sets of candidates A 2 = the events of second sets of candidates B= the event of introducing a new product. We are given, P (A 1 )= 0.65 P (A 2 )= 0.35 P (B/A 1 ) = 0.80 P (B/A 2 ) = 0.30 Probability of introducing the product = P(B) = P (A 1 and B) + P (A 2 and B)
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23 2 2 1 1 A B P ) (A P A B P ) P(A (B) P = 0.65 0.80 + 0.35 0.30 = 0.52 + 0.105 = 0.625. Illustration 2 A factory has two machines. Past records show that the first machine produces 40 per cent of output and the second machine produces 60 per cent of output. Further, 4 per cent and 2 per cent of products produced by the first machine and the second machine were defectives. If a defective item is drawn at random, what is the probability that the defective item was produced by the first machine or the second machine. Solution A 1 = the event of drawing an item produced by the first machine A 2 = the event of drawing an item produced by the second machine B = the event of drawing defective item produced by either first machine or second Machine. Based on the information given, we can have, P ( A 1 )= 100 40 = 0.40 , P (A 2 )= 100 60 = 0.60 0.04 100 4 A B P 1 0.02 100 2 A B P 2 P(B) = P (A 1 and B) + P (A 2 and B) P(B) 2 2 1 1 A B P ) (A P A B P ) P(A = 0.40 0.04 + 0.60 0.02 = 0.028 (B) P ) (B/A P ) P(A (B) P B) and P(A B A P 1 1 1 1 0.571 0.028 0.04 0.40 (B) P ) (B/A P ) P(A (B) P B) and P(A B A P 2 2 2 2 0.429 0.028 0.02 0.60
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24 The values derived above can also compute in tabular form, as below. Computation of posterior probabilities. Event (1) Prior Probability P(A 1 ) (2) Conditional probability P (B/A 1 ) (3) Joint probability P (A 1 and B) (4) Posterior Probability P (A 1 /B), 4-P(B) (5) A 1 A 2 0.40 0.60 0.04 0.02 0.016 0.012 012 . 0 0.028 0.016 028 . 0 012 . 0 = 0.5714 = 0.4286 1.00 P(B) = 0.028 1.0000 Conclusion With additional information i.e. the probability of defective items produced by the first machine is 0.5714 or 57.14 per cent and that by the second machine is 0.4286 or 42.86 per cent. And we may say that the defective item is more likely drawn from the output produced by the machine. 4.4. REVISION POINTS ) A ( P ) AB ( P A B P ) B ( P ) AB ( P B A P In conditional probabilities, the rule of multiplication in its modified form is : P (A and B) = P (A) P A B P (A and B) = P (B) P B A 4.5. INTEXT QUESTION Suppose that one of the three men, a politician, a businessman and an academician will be appointed as the Vice-Chancellor of a University. Their probability appointments respectively are 0.40, 0.25 and 0.35. the probabilities that
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