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Visualizing Events

Visualizing Events. Contingency Tables Tree Diagrams. Ace Not Ace Total. Black 2 24 26. Red 2 24 26.

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Visualizing Events

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

  2. Contingency Table A Deck of 52 Cards Red Ace Not an Ace Total Ace Red 2 24 26 Black 2 24 26 Total 4 48 52 Sample Space

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

  4. Probability • Probabilityis the numerical • measure of the likelihood • that the event will occur. • Value is between 0and 1. • Sumof the probabilities of • all mutually exclusive • events is1. 1 Certain .5 0 Impossible

  5. 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)

  6. Joint Probability Using Contingency Table Event Total B1 B2 Event A1 P(A1andB1) P(A1andB2) P(A1) A2 P(A2 and B1) P(A2andB2) P(A2) 1 Total P(B1) P(B2) Marginal (Simple) Probability Joint Probability

  7. Addition Rule P(A1or B1 ) = P(A1) +P(B1) - P(A1 andB1) Event Total B1 B2 Event A1 P(A1and B1) P(A1and B2) P(A1) P(A2 andB1) A2 P(A2 and B2) P(A2) 1 Total P(B1) P(B2) For Mutually Exclusive Events: P(A or B) = P(A) + P(B)

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

  9. Computing Conditional Probability The Probability of the Event: EventAgiventhat EventBhas occurred P(A êB) = e.g. P(Red Card given that it is anAce) =

  10. Conditional Probability Using Contingency Table Conditional Event: Draw 1 Card. Note Kind & Color Color Type Total Red Black Revised Sample Space 2 2 4 Ace 24 24 48 Non-Ace 26 26 52 Total

  11. Conditional Probability and Statistical Independence Conditional Probability: P(AçB) = P(A and B) = P(AêB) • P(B) Multiplication Rule: Events are Independent: P(AêB) = 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.

  12. Bayes’ Theorem:Contingency Table What are the chances of repaying a loan, given a college education? Loan Status Prob. Education Repay Default .2 .05 .25 College ? ? ? No College ? ? 1 Prob. ê P(RepayCollege) =

  13. Discrete Probability Distribution • List ofAll Possible[ Xi, P(Xi) ]Pairs • Xi = Value of Random Variable (Outcome) • P(Xi) = Probability Associated with Value • Mutually Exclusive(No Overlap) • Collectively Exhaustive(Nothing Left Out) • 0 £P(Xi) £ 1 • SP(Xi) = 1

  14. 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 Probabilityfor 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

  15. 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

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

  17. Binomial Distribution Characteristics n = 5p = 0.1 P(X) Mean .6 = E ( X ) = np m .4 .2 e.g. m=5 (.1) =.5 0 X 0 1 2 3 4 5 Standard Deviation n = 5p = 0.5 P(X) ) s = - p np ( 1 .6 .4 .2 e.g.s=5(.5)(1 - .5)= 1.118 X 0 0 1 2 3 4 5

  18. P ( X = x | l - l x e l x ! Poisson Distribution • Poisson Process: • Discrete Eventsin an ‘Interval’ • The Probability ofOne Successin Interval is Stable • The Probability of More than One Success in this Interval is 0 • Probability of Success is • Independentfrom Interval to • Interval • e.g. # Customers Arriving in 15 min. • # Defects Per Case of Light Bulbs.

  19. Poisson Probability Distribution Function -l X e l P ( X ) = X ! P(X ) = probability ofXsuccesses givenl l= expected (mean) number of ‘successes’ e = 2.71828 (base of natural logs) X = number of ‘successes’per unit e.g. Find the probability of4 customers arriving in 3 minutes when the mean is3.6. -3.6 4 e 3.6 P(X)= =.1912 4!

  20. Poisson Distribution Characteristics Mean l = 0.5 P(X) .6 m = E ( X ) = l .4 N .2 å = 0 X X P ( X ) i i 0 1 2 3 4 5 i = 1 l = 6 P(X) .6 Standard Deviation .4 .2 s = l 0 X 0 2 4 6 8 10

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