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= P (B1 ï A) ï (B2 ï A) ï . . . ï (Bn ï A) = n ï¥ P( B i i ï½1 = ï A) n ï¥ P( B ) P( A \ B ) i i ï½1 from Multiplication Rule i Bayesâ Rule If the mutually exclusive events B1 , B2 , . .. , Bn constitute a partition of the sample space S such that P(Bi) ï¹ 0 for i = 1, 2, â¦, n, then for any event A in S such that P(A) ï¹ 0 P (Bj \A) = P( B j ï A) P( B j ï A) = P( A) n ï¥ P( B i i ï½1 = ï A) = P( B j ï A) n ï¥ P( B ) P( A \ B ) i ï½1 i i P( A \ B j ) P( B j ) n ï¥ P( A \ B ) P( B ) i ï½1 P (Bj \A) = i i P( A \ B j ) P( B j ) n ï¥ P( A \ B ) P( B ) i ï½1 i Bayesâ Rule i Bayesâ Rule is an extremely useful formula. It can be used to find the conditional probability P(Bj\A) when the available information cannot be used to apply the definition of conditional probability directly. Ex. 19 A factory manufactures products by using 4 machines which can produce 1000, 1200, 1800, and 2000 pieces a day, respectively. The machines produce defective products at the rates of 1 %, ½ %, ½ %, and 1 %, respectively. A product is selected from a lot which is produced in one day and found to be defective. Determine the probability that this product is manufactured by the 4th machine.
