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 anAce

Total

Red

Black

Total

2 24

2 24

26

26

4 48 52

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.

SpaceSampleinOutcomesTotal

OutcomesEventofNumber)E(P

T

X

e.g. P( ) = 2/36

(There are 2 ways to get one 6 and the other 4)

P(A2 and B2)

P(A1 and B2)

P(A2 and B1)

P(A1 and B1)

EventEvent Total

Total 1

Joint Probability Using Contingency Table

Joint Probability Marginal (Simple) Probability

P(A1)A1

A2

B1 B2

P(A2)

P(B1) P(B2)

P(A1 and B1)

P(B2)P(B1)

P(A2 and B2)P(A2 and B1)

EventEvent Total

Total 1

Addition Rule

P(A1 and B2) P(A1)A1

A2

B1 B2

P(A2)

P(A1 or B1 ) = P(A1) +P(B1) - P(A1 and B1)

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 FaceLight 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) =

)B(P

)BandA(P

2

1

4

2

Aces

AcesdRe

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

Conditional Probability Using Contingency Table

Conditional Event: Draw 1 Card. Note Kind & Color

26

2

5226

522

/

/

P(Red)

Red)AND P(Ace = Red) |P(Ace

Revised Sample Space

Conditional Probability and Statistical Independence

)B(P

)BandA(PConditional Probability: P(AB) =

P(A and B) = P(A B) • P(B)

Events areIndependent:

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 StatusEducation Repay Default Prob.

College .2 .05 .25

No College

Prob. 1

P(Repay College) = 08.

)DefaultandCollege(P)payReandCollege(P

)payReandCollege(P

? ? ?

? ?

Discrete Probability Distribution

• List of All 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

• P(Xi) = 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 ! n Xp pX 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

E X np

np p

( )

( )1

0.2.4.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

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.

P X x

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

P XX

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

3.64!

4

= .1912

Poisson Distribution Characteristics

= 0.5

= 6

Mean

Standard Deviation

ii

N

i

E X

X P X

( )

( )1

0.2.4.6

0 1 2 3 4 5

X

P(X)

0.2.4.6

0 2 4 6 8 10

X

P(X)