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Visualizing Events Contingency Tables Tree Diagrams Ace Not Ace Total Red 2 24 26 Black 2 24 26 Total 4 48 52

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Contingency Table A Deck of 52 Cards Ace Not an Ace Total Red Black Total 224 2 26 44852 Sample Space Red Ace

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Tree Diagram Event Possibilities Red Cards Black Cards Ace Not an Ace Ace Not an Ace Full Deck of Cards

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Probability Probability is the numerical measure of the likelihood that the event will occur. Value is between 0 and 1. Sum of the probabilities of all mutually exclusive events is 1. Certain Impossible.5 1 0

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Computing Probability The Probability of an Event, E: Each of the Outcome in the Sample Space equally likely to occur. e.g. P ( ) = 2/36 (There are 2 ways to get one 6 and the other 4)

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P(A 2 and B 2 ) P(A 1 and B 2 ) P(A 2 and B 1 ) P(A 1 and B 1 ) Event Total 1 Joint Probability Using Contingency Table Joint Probability Marginal (Simple) Probability P(A 1 )A1A1 A2A2 B1B1 B2B2 P(A 2 ) P(B 1 )P(B 2 )

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P(A 1 and B 1 ) P(B 2 )P(B1)P(B1) P(A 2 and B 2 ) P(A 2 and B 1 ) Event Total 1 Addition Rule P(A 1 and B 2 )P(A 1 )A1A1 A2A2 B1B1 B2B2 P(A 2 ) P(A 1 or B 1 ) = P(A 1 ) +P(B 1 ) - P(A 1 and B 1 ) For Mutually Exclusive Events: P(A or B) = P(A) + P(B)

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3 Items, 3 Happy Faces Given they are Light Colored Dependent or Independent Events The Event of a Happy Face GIVEN it is Light Colored E = Happy Face Light Color

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Computing Conditional Probability The Probability of the Event: Event A given that Event B has occurred P(A B) = e.g. P(Red Card given that it is an Ace) =

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Black Color Type Red Total Ace 224 Non-Ace 24 48 Total 26 52 Conditional Probability Using Contingency Table Conditional Event: Draw 1 Card. Note Kind & Color Revised Sample Space

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Conditional Probability and Statistical Independence Conditional Probability: P(A B) = P(A and B) = P(A B) P(B) Events are Independent: P(A B) = P(A) Or, P(A and B) = P(A) P(B) Events A and B are Independent when the probability of one event, A is not affected by another event, B. Multiplication Rule:

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What are the chances of repaying a loan, given a college education? Bayes’ Theorem: Contingency Table Loan Status Education RepayDefault Prob. College.2.05.25 No College Prob. 1 P(Repay College) = ? ?? ? ?

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Discrete Probability Distribution List of All Possible [ X i, P(X i ) ] Pairs X i = Value of Random Variable (Outcome) P(X i ) = Probability Associated with Value Mutually Exclusive (No Overlap) Collectively Exhaustive (Nothing Left Out) 0 P(X i ) 1 P(X i ) = 1

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Binomial Probability Distributions ‘n’ Identical Trials, e.g. 15 tosses of a coin, 10 light bulbs taken from a warehouse 2 Mutually Exclusive Outcomes, e.g. heads or tails in each toss of a coin, defective or not defective light bulbs Constant Probability for each Trial, e.g. probability of getting a tail is the same each time we toss the coin and each light bulb has the same probability of being defective

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Binomial Probability Distributions 2 Sampling Methods: Infinite Population Without Replacement Finite Population With Replacement Trials are Independent: The Outcome of One Trial Does Not Affect the Outcome of Another

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Binomial Probability Distribution Function P(X) = probability that X successes given a knowledge of n and p X = number of ‘successes’ in sample, (X = 0, 1, 2,..., n) p = probability of ‘success’ n = sample size P(X) n X ! nX pp X n X ! ()! ( ) 1 Tails in 2 Toss of Coin X P(X) 0 1/4 =.25 1 2/4 =.50 2 1/4 =.25

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Binomial Distribution Characteristics n = 5 p = 0.1 n = 5 p = 0.5 Mean Standard Deviation EXnpnp npnp p () ( ) 1 0.2.4.6 012345 X P(X).2.4.6 012345 X P(X) e.g. = 5 (.1) =.5 e.g. = 5(.5)(1 -.5) = 1.118 0

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Poisson Distribution Poisson Process: Discrete Events in an ‘Interval’ – The Probability of One Success in Interval is Stable – The Probability of More than One Success in this Interval is 0 Probability of Success is Independent from Interval to Interval e.g. # Customers Arriving in 15 min. # Defects Per Case of Light Bulbs. PXx x x (| ! e -

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Poisson Probability Distribution Function P(X ) = probability of X successes given =expected (mean) number of ‘successes’ e=2.71828 (base of natural logs) X=number of ‘successes’ per unit PX X X () ! e e.g. Find the probability of 4 customers arriving in 3 minutes when the mean is 3.6. P(X) = e -3.6 3.6 4! 4 =.1912

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Poisson Distribution Characteristics = 0.5 = 6 Mean Standard Deviation i i N i EX XPX () () 1 0.2.4.6 012345 X P(X) 0.2.4.6 0246810 X P(X)

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