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1 Probability- Independence and Fundamental Rules Dr. Jerrell T. Stracener, SAE Fellow EMIS 7370 STAT 5340 Probability and Statistics for Scientists and Engineers Leadership in Engineering Department of Engineering Management, Information and Systems

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2 Two events A and B in S are independent if, and only if, P(A B) = P(A)P(B) Definition - Independence

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3 If A, B and C are independent events, in S, then P(A B C) = P(A)P(B)P(C) If A 1, …, A n are independent events in S, then Rules of Probability

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4 Mutually Exclusive Events If A and B are any two events in S, then P(A B) = 0 Complementary Events If A' is the complement of A, then P(A') = 1 - P(A) Rules of Probability

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5 Rule: If A and B are any two events in S, then P(A B) = P(A) + P(B) - P(A B) Rule: If A and B are mutually exclusive, then P(A B) = P(A) + P(B) Rules of Probability - Addition Rules

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6 Rule For any 3 events, A, B and C in S, P(A B C) = P(A) + P(B) + P(C) - P(A B) - P(A C) - P(B C) + P(A B C) Rule If A, B and C are mutually exclusive events in S, then P(A B C) = P(A) + P(B) + P(C) Rules of Probability

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Example: Coin Tossing Game A game is played as follows: a)A player tosses a coin two times in sequence. If at least one head occurs the player wins. What is the probability of winning? a)The game is modified as follows: A player tosses a coin. If a head occurs, the player wins. Otherwise, the coin is tossed again. If a head occurs, the player wins. What is the probability of winning?

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8 Example An biased coin (likelihood of a head is 0.75) is tossed three times in sequence. What is the probability that 2 heads will occur?

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9 If A 1, A 2,..., A n are mutually exclusive, then P(A 1 A 2 ... A n ) = P(A 1 ) + P(A 2 ) P(A n ) Rule For events A 1, A 2,..., A n, Rules of Probability continued

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10 A town has two fire engines operating independently. The probability that a specific fire engine is available when needed is (a) What is the probability that neither is available when needed? (b) What is the probability that a fire engine is available when needed? Exercise - Fire Engines

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11 a. An electrical circuit consists of 4 switches in series. Assume that the operations of the 4 switches are statistically independent. If for each switch, the probability of failure (i.e., remaining open) is 0.02, what is the probability of circuit failure? b. Rework for the case of 4 switches in parallel. Exercise - 4 Switches in Series

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12 Series configuration P(switch fails) = 0.02 P(switch does not fail) = = 0.98 By observing the diagram above, the circuit fails if at least one switch fails. Also the circuit is a success if all 4 switches operate successfully. S1S1 S2S2 S3S3 S4S4 Exercise - 4 Switches in Series - solution

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13 Therefore, P(circuit failure) = 1 - P(circuit success) = 1 - P(S 1 S 2 S 3 S 4 ) = 1 - (0.98) 4 = Exercise - 4 Switches in Series - solution

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14 Parallel configuration P(circuit failure) = P(4 of 4 switches fail) = P(S 1 S 2 S 3 S 4 ) = P(S 1 )P( S 2 )P( S 3 )P( S 4 ) = (0.02) 4 = S1S1 S2S2 S3S3 S4S4 Exercise - 4 Switches in Series - solution

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15 A rental car service facility has 10 foreign cars and 15 domestic cars waiting to be serviced on a particular Saturday morning. Because there are so few mechanics working on Saturday, only 6 can be serviced. If the 6 are chosen at random, what is the probability that at least 3 of the cars selected are domestic? Example - Car Rental

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16 Let A = exactly 3 of the 6 cars chosen are domestic. Assuming that any particular set of 6 cars is as likely to be chosen as is any other set of 6, we have equally likely outcomes, so where n is the number of ways of choosing 6 cars from the 25 and n A is the number of ways of choosing 3 domestic cars and 3 foreign cars. Thus To obtain n A, think of first choosing 3 of the 15 domestic cars n Example - Car Rental - solution

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17 and then 3 of the foreign cars. There are ways of choosing the 3 domestic cars, and there are ways of choosing the 3 foreign cars; n A is the product of these two numbers (visualize a tree diagram) using a product rule, so Example - Car Rental - solution

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18 Let D 4 = (exactly 4 of the 6 cars chosen are domestic), and define D 5 and D 6 in an analogous manner. Then the probability that at least 3 domestic cars are selected is P(D 3 D 4 D 5 D 6 ) = P(D 3 ) + P(D 4 ) + P(D 5 ) + P(D 6 ) This is also the probability that at most 3 foreign cars are selected. Example - Car Rental - solution

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