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

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

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

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

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)

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 )

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)

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

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) =

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

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:

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) = ? ?? ? ? 

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

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

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

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

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

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 -

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

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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