P a c p b a c and shows that for any event a with 0 p

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P A c   P B | A c and shows that, for any event A with 0 P A 1, we can write P B as a weighted average of the two conditional probabilities P B | A and P B | A c , with respective weights P A and P A c 1 P A . 63
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Except by fluke, it is not true that P B | A P B | A c 1. [By contrast, it is true that P B | A P B c | A 1 always.] To get Bayes’ rule, we use the previous expression for P B : P A | B P B | A   P A P B | A   P A P B | A c   P A c In other words, we can obtain P A | B by knowing P B | A , P B | A c , and P A . 64
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EXAMPLE : Deciding whether to locate a business at a particular site. Want to know the chances that the business will be profitable. Can conduct a study that provides input. A site is profitable B study says site is profitable 65
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Suppose 32% of all similar sites are profitable, so P A .32. Based on past experience where studies were conducted, also know P B | A .65 and P B | A c .29. Using Bayes’ rule, P A | B .65  .32 .65  .32 .29  .68 .513 In other words, if the study says the site will be profitable, the site actually will be profitable with probability .513. This is notably higher than the unconditional probability – that is, the probability without knowing the outcome of the study – which is .32. What is P A | B c ? 66
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