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PROBABILITY

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Experiments With Uncertain Outcomes Experiment Toss a coin Roll a die Inspect a part Conduct a survey Hire New Employees Find Errors on Tax Form Complete a Task Weigh a ContainerOutcomes Heads/Tails 1, 2, 3, 4, 5, 6 Defective/OK Yes/No 0, 1, 2, days 0 – 25 pounds

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Simple Events and Events Simple Event –One of the possible outcomes (that cannot be further broken down) Sample Space –Set of all possible simple events Mutually Exclusive Exhaustive Event –A collection of one or more simple events

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PROBABILITY CONCEPTS ProbabilityProbability –The likelihood an event will occur Basic Requirements for Assigning ProbabilitiesBasic Requirements for Assigning Probabilities 1.The probability of all events lies between 0 and 1 2.The sum of the probabilities of all simple events = 1

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3 Approaches to Assigning Probabilities A priori Classical Approach –Games of chance Relative Frequency Approach –Long run likelihood of an event occurring Subjective Approach –Best estimates

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Classical Approach Assume there are N possible outcomes of an experiment and they are all equally likely to occur Assigning Probability –Suppose X of the outcomes correspond to the event A. Then the probability that event A will occur, written P(A) is: P(A) = X/N Example: P(Club) = # clubs/52 = 13/52

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Relative Frequency Approach Long term behavior of an event A has been observed n observations P(A) = (#times A occurred) / n Example: n = 800 students take statistics 164 received an A P(Receiving an A) = 164/800

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Subjective Approach These are best estimate probabilities based on experience and knowledge of the subject Example: A meteorologist uses charts of wind flow and pressure patterns to predict that the P(it will rain tomorrow ) =.75 –This will be stated as a 75% chance of rain tomorrow

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PROBABILITIES OF COMBINATIONS OF EVENTS Joint Probability P(A and B) = Probability A and B will occur simultaneously Marginal Probability P(A) = (Probabilities of all the simple events that contain A) Either/Or Probability -- Addition Rule P(A or B) = P(A) + P(B) - P(A and B) Conditional Probability P(A|B) = P(A and B)/P(B) Joint Probability (Revisited) P(A and B) = P(A|B)P(B) = P(B|A)P(A) Complement Probability

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INDEPENDENCE Events A and B are independent if knowing B does not affect the probability that A occurs or vice versa, i.e. P(A|B) = P(A) and P(B|A) = P(B) IndependentJoint Probability (For Independent Events) P(A and B) = P(A)P(B|A) = P(A)P(B) if A and B are independent A Test for IndependenceA Test for Independence -- Check to see if: P(A and B) = P(A)P(B) If it does =====> Independent If not, =====> Dependent

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Mutually Exclusive and Exhaustive Events Events A and B are mutually exclusive if: P(A and B) = 0 –Thus if A and B are mutually exclusive, P(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) Events A, B, C, D are exhaustive if: P(at least one of these occurs) = 1

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Example 200 people from LA, OC and SD surveyed: Do you favor gun control? YES NO ? LA OC SD No opinion

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Example: P(LA) = P(LA and Yes) + P(LA and NO) + P(LA and ?) = =.40 Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 Joint Probability Table YES NO ? LA OC SD JOINT PROB. P(YES and LA) =40/200

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What is the probability a randomly selected person is from LA and favors gun control? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD –P(LA and Yes) =.20 (from table)

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What is the probability a randomly selected person is opposed to gun control? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD –P(NO) =.35 (in the margin of the table)

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not What is the probability a randomly selected person is not from San Diego? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD = =

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Joe is from LA. What is the probability Joe favors gun control? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD P(YES|LA) = P(YES and LA)/P(LA) We know (we are given that) Joe is from LA. =.20/.40 =.50

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Bill is opposed to gun control. What is the probability Bill is from Orange County? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD We know (we are given that) Bill is opposed to gun control. P(OC|NO) = P(OC and NO)/P(NO) =.05/.35 =.143

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What is the probability that a randomly selected person is from LA or favors gun control? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD P(LA or Yes) = P(LA) + P(YES) - P(LA and YES) =.70 =

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Are being from San Diego and having no opinion on gun control a pair of mutually exclusive events? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD Does P(SD and ?) = 0 are They are mutually exclusive. YES

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Are being from Orange County and having no opinion on gun control a pair of mutually exclusive events? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD Does P(OC and ?) = 0 are not They are not mutually exclusive. NO

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Are being from LA and favoring gun control a pair of independent events? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD Does P(LA and YES) = P(LA)P(YES)? =.20 =.20 YES LA and YESare independent = (.40) (.50)?

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Are being from San Diego and favoring gun control a pair of independent events? Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD Does P(SD and YES) = P(SD)P(YES)? =.10 =.10 NO SD and YES are not independent = (.20) (.50)?

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Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD Are being from LA, being from Orange County, favoring gun control, and opposing gun control form a set of exhaustive events? No positive probability remains Events are exhaustive

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Marginal.40 = P(LA).40 = P(OC).20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES NO ? LA OC SD Are being from Orange County, being from San Diego, favoring gun control, and opposing gun control form a set of exhaustive events? There is positive probability remaining. Events are not exhaustive

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Calculating Probabilities Using Venn Diagrams Convenient way of depicting some of the logical relationships between events Circles can be used to represent events –Overlapping circles imply joint events –Circles which do not overlap represent mutually exclusive events –The area outside a region is the complement of the event represented by the region

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Example Students at a college have either Microsoft Explorer (E), Netscape (N), both or neither browsers installed on their home computers P(E) =.85 and P(N) =.50 P(both) =.45 What is the probability a student has neither? Neither E (.85) P(E or N) = =.90 P(neither) = =.10 N (.50) (.45) E and N

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Probability Trees Probability Trees are a convenient way of representing compound events based on conditional probabilities –They express the probabilities of a chronological sequence of events Example: The probability of winning a contract is.7. If you win the contract P(hiring new workers) =.8 If you do not win the contract P(hiring new workers) =.4 What is the probability you will hire new workers?

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The Probability Tree –Start with whether or not you win the contract –Then for each possibility list the probability of hiring new workers –Multiply the probabilities and add appropriate ones Win contract (.7) Lose Contract (.3) Hire new workers (.8) Do not hire new workers (.2) Hire new workers (.4) Do not hire new workers (.6)

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REVIEW Probabilities are measures of likelihood How to determine probabilities Joint, marginal, conditional probabilities Complement and “either/or” probabilities Mutually exclusive, independent and exhaustive events Venn diagrams Decision trees

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