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Unformatted text preview: 2. .15 + .68 .08 = .75 3. P(Type B computer) = 120 200 = .60, so P(Type A computer) = .40 There are 120 Type B computers, and 75 of them do not experience any problems, so 45 of them do. That makes 95 computers that experience problems. Therefore, P(Type A computer experience problems) = 50 95 = .526 .40. So the two events are not independent. 2 4. Let A,B,C denote the events that the plug is type A,B,C respectively, and D the event that the plug is defective. We have...
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- Summer '07