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PROBABILITY. 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 Container. Outcomes Heads/Tails 1, 2, 3, 4, 5, 6 Defective/OK Yes/No 0, 1, 2 , 3 0 - 64 0-10 days 0 – 25 pounds.

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## PROBABILITY

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**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 Container Outcomes Heads/Tails 1, 2, 3, 4, 5, 6 Defective/OK Yes/No 0, 1, 2 , 3 0 - 64 0-10 days 0 – 25 pounds Experiments WithUncertain Outcomes**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**PROBABILITY CONCEPTS**• Probability • The likelihood an event will occur • Basic Requirements for Assigning Probabilities • The probability of all events lies between 0 and 1 • The sum of the probabilities of all simple events = 1**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**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**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**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**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**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) • Joint 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 Independence -- Check to see if: P(A and B) = P(A)P(B) If it does =====> Independent If not, =====> Dependent**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**No opinion**Example 200 people from LA, OC and SD surveyed: Do you favor gun control? YES NO ? LA 40 30 10 OC 50 10 20 SD 10 30 0**JOINT PROB.**P(YES and LA) =40/200 Joint Probability Table Example: P(LA) = P(LA and Yes) + P(LA and NO) + P(LA and ?) = .20 + .15 + .05 = .40 YES NO ? LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 • P(LA and Yes) = .20 (from table) What is the probability a randomly selected person is from LA and favors gun control?**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 • P(NO) = .35 (in the margin of the table) What is the probability a randomly selected person is opposed to gun control?**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 = 1-.20 = What is the probability a randomly selected person is not from San Diego? .80**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 = .20/.40 Joe is from LA. What is the probability Joe favors gun control? We know (we are given that) Joe is from LA. = .50 P(YES|LA) = P(YES and LA)/P(LA)**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 = .05/.35 Bill is opposed to gun control. What is the probability Bill is from Orange County? We know (we are given that) Bill is opposed to gun control. = .143 P(OC|NO) = P(OC and NO)/P(NO)**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 = .40 + .50 - .20 What is the probability that a randomly selected person is from LA or favors gun control? P(LA or Yes) = P(LA) + P(YES) - P(LA and YES) =.70**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 YES Are being from San Diego and having no opinion on gun control a pair of mutually exclusive events? Does P(SD and ?) = 0 They are mutually exclusive.**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 NO Are being from Orange County and having no opinion on gun control a pair of mutually exclusive events? Does P(OC and ?) = 0 They are not mutually exclusive.**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 .20 = (.40) • (.50)? Are being from LA and favoring gun control a pair of independent events? LA and YES are independent Does P(LA and YES) = P(LA)•P(YES)? = .20 YES**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 .05 = (.20) • (.50)? Are being from San Diego and favoring gun control a pair of independent events? SD and YES are not independent Does P(SD and YES) = P(SD)•P(YES)? = .10 NO**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) No positive probability remains Events are exhaustive Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 Are being from LA, being from Orange County, favoring gun control, and opposing gun control form a set of exhaustive events?**YES NO ?**LA .20 .15 .05 OC .25 .05 .10 SD .05 .15 0 Marginal .40 = P(LA) .40 = P(OC) .20 = P(SD) There is positive probability remaining. Events are not exhaustive Marginal P(YES) P(NO) P(?) =.50 =.35 =.15 Are being from Orange County, being from San Diego, favoring gun control, and opposing gun control form a set of exhaustive events?**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**E and N**(.45) N E (.85) (.50) Neither 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? P(E or N) = .85+.50-.45= .90 P(neither) = 1-.90 = .10**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?**Hire new workers (.8)**Win contract (.7) .68 Do not hire new workers (.2) Hire new workers (.4) Lose Contract (.3) Do not hire new workers (.6) 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 .56 .14 .12 .18**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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