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Chapter 4: Probability (Cont.)

Chapter 4: Probability (Cont.). In this handout: Venn diagrams Event relations Laws of probability Conditional probability Independence of events. Venn Diagram: representing events graphically.

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Chapter 4: Probability (Cont.)

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  1. Chapter 4: Probability (Cont.) • In this handout: • Venn diagrams • Event relations • Laws of probability • Conditional probability • Independence of events

  2. Venn Diagram: representing events graphically Example: Toss a coin twice. Let event A corresponds to “ tail at the second toss ”; event B corresponds to “ at least one head ”.

  3. Example: Two monkeys to be selected by lottery for an experiment. Label the possible pairs (elementary outcomes):{1, 2} e1 {2, 3} e4{1, 3} e2 {2, 4} e5{1, 4} e3 {3, 4} e6 Let A: selected monkeys are of the same type;B: selected monkeys are of the same age;

  4. The complement of an event A is the set of all elementary outcomes that are not in A.The union of events A and B isthe set of all elementary outcomes that are in A, B, or both.The intersection of events A and B is the set of all elementary outcomes that are in A and B.

  5. Example: Equal chances that the answer to a problem is correct or wrong. What is the probability of getting at least one correct answer? P(at least one correct answer) = 1 – P(all answers wrong) = 1 – 1/8 = 7/8

  6. Conditional probability The probability of an event A must often be modified after information is obtained as to whether or not a related event B has taken place.Example: Q1: Probability that a randomly-selected person has hypertension? Q2: A randomly-selected person is overweight. What is the probability that the person also has hypertension? Let A denote “has hypertension”, B denote “overweight”. Then P(has hypertension given that overweight) = P( A | B ) = .1/.25 = .4

  7. Conditional probability Box on Page 143Conditioned probability; multiplication law of probability

  8. Independence of Events

  9. Independence of Events Example: A mechanical system consists of two components. Component 1 has reliability (probability of not failing) .98 and component 2 has reliability .95. If the system can function only if both components function, what is the reliability of the system? Let A1 denote “component 1 functions”, A2 denote “component 2 functions”, S denote “system functions”. Given that the components operate independently, we take the events A1 and A2 to be independent. Thus, P(S) = P(A1) P(A2) = .98 * .95 = .931

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