PROBABILITY

PROBABILITYPROBABILITY

Experiments WithUncertain Outcomes

ExperimentExperiment• 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

OutcomesOutcomes• Heads/Tails• 1, 2, 3, 4, 5, 6• Defective/OK• Yes/No• 0, 1, 2 , 3• 0 - 64• 0-10 days • 0 – 25 pounds

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

• ProbabilityProbability– The likelihood an event will occur

• Basic Requirements for Assigning ProbabilitiesBasic Requirements for Assigning Probabilities1. The probability of all events lies between 0 and 1

2. 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 EVENTSJoint Probability

P(A and B) = Probability A and B will occur simultaneously

Marginal ProbabilityP(A) = (Probabilities of all the simple events that contain A)

Either/Or Probability -- Addition RuleP(A or B) = P(A) + P(B) - P(A and B)

Conditional ProbabilityP(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 ProbabilityP(A)1)AP(

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 IndependentIndependent Events) P(A and B) = P(A)P(B|A) = P(A)P(B) if A and B are independent

• A Test for IndependenceA 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

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

No opinion

Example:

P(LA) = P(LA and Yes) + P(LA and NO) + P(LA and ?)

= .20 + .15 + .05 = .40

Marginal.40 = P(LA).40 = P(OC).20 = P(SD)

Marginal P(YES) P(NO) P(?) =.50 =.35 =.15

Joint Probability Table

YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0JOINT PROB.P(YES and LA)

=40/200

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

–P(LA and Yes) = .20 (from 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

–P(NO) = .35 (in the margin of the table)

What is the probability a randomly selected person is notnot from San Diego?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

P(SD)1)SDP( .80.80= 1-.20 =

Joe is from LA. What is the probability Joe favors gun control?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

P(YES|LA) = P(YES and LA)/P(LA)

We know (we are given that) Joe is from LA.

= .20/.40 = .50

Bill is opposed to gun control. What is the probability Bill is from Orange County?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

We know (we are given that) Bill is opposed to gun control.

P(OC|NO) = P(OC and NO)/P(NO) = .05/.35 = .143

What is the probability that a randomly selected person is from LA or favors gun

control?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

P(LA or Yes) = P(LA) + P(YES) - P(LA and YES) =.70 = .40 + .50 - .20

Are being from San Diego and having no opinion on gun control a pair of mutually

exclusive events?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

Does P(SD and ?) = 0

They areare mutually exclusive.

YES

Are being from Orange County and having no opinion on gun control a pair

of mutually exclusive events?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

Does P(OC and ?) = 0

They are notare not mutually exclusive.

NO

Are being from LA and favoring gun control a pair of independent events?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

Does P(LA and YES) = P(LA)•P(YES)?

= .20 = .20 YESYES

LA and YES

areare

independent .20.20 = (.40) • (.50)?

Are being from San Diego and favoring gun control a pair of independent events?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

Does P(SD and YES) = P(SD)•P(YES)?

= .10 = .10 NONO

SD and YES

are notare not

independent .05.05 = (.20) • (.50)?



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

Are being from LA, being from Orange County, favoring gun control, and opposing gun control form a set of exhaustive events?

No positiveprobability remains

Events are exhaustive



YES NO ? LA .20 .15 .05 OC .25 .05 .10

SD .05 .15 0

Are being from Orange County, being from San Diego, favoring gun control, and opposing

gun control form a set of exhaustive events?

There is positiveprobability remaining.

Events are not exhaustive

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

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?

NeitherEE

(.85)(.85)

P(E or N) =

.85+.50-.45=

.90P(neither) =

1-.90 = .10

NN(.50)(.50)

(.45)(.45)E and N

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?

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

Win contract (.7)

Lose Contract (.3)

Hire new workers (.8)

Do not hire new workers (.2)

Hire new workers (.4)

Do not hire new workers (.6)

.14

.56

.12

.18

.68

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

Documents

PROBABILITY