Chapter 4Introduction to Probability

� Experiments, Counting Rules, and Assigning Probabilities

� Events and Their Probability� Some Basic Relationships

of Probability� Conditional Probability

Probability as a Numerical Measureof the Likelihood of Occurrence

0 1.5

Increasing Likelihood of Occurrence

Probability:

The eventis very

unlikelyto occur.

The occurrenceof the event isjust as likely asit is unlikely.

The eventis almostcertain

to occur.

An Experiment and Its Sample Space

An experiment is any process that generateswell-defined outcomes.

The sample space for an experiment is the set ofall experimental outcomes.

An experimental outcome is also called a samplepoint.

Example: Bradley Investments

Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that thepossible outcomes of these investments three monthsfrom now are as follows.

Investment Gain or Lossin 3 Months (in $000)

Markley Oil Collins Mining1050

−20

8−2

A Counting Rule for Multiple-Step Experiments

If an experiment consists of a sequence of k stepsin which there are n1 possible results for the first step,n2 possible results for the second step, and so on, then the total number of experimental outcomes isgiven by (n1)(n2) . . . (nk).

A helpful graphical representation of a multiple-stepexperiment is a tree diagram.

Bradley Investments can be viewed as atwo-step experiment. It involves two stocks, eachwith a set of experimental outcomes.

Markley Oil: n1 = 4Collins Mining: n2 = 2Total Number of Experimental Outcomes: n1n2 = (4)(2) = 8

A Counting Rule for Multiple-Step Experiments

Tree Diagram

Gain 5Gain 5

Gain 8Gain 8

Gain 8Gain 8

Gain 10Gain 10

Gain 8Gain 8

Gain 8Gain 8

Lose 20Lose 20

Lose 2Lose 2

Lose 2Lose 2

Lose 2Lose 2

Lose 2Lose 2

EvenEven

Markley Oil(Stage 1)

Collins Mining(Stage 2)

ExperimentalOutcomes

(10, 8) Gain $18,000

(10, -2) Gain $8,000

(5, 8) Gain $13,000

(5, -2) Gain $3,000

(0, 8) Gain $8,000

(0, -2) Lose $2,000

(-20, 8) Lose $12,000

(-20, -2) Lose $22,000

A second useful counting rule enables us to count thenumber of experimental outcomes when n objects are tobe selected from a set of N objects.

Counting Rule for Combinations

CNn

Nn N nn

N =⎛⎝⎜

⎞⎠⎟

=−!

!( )!C

Nn

Nn N nn

N =⎛⎝⎜

⎞⎠⎟

=−!

!( )!

Number of Combinations of N Objects Taken n at a Time

where: N! = N(N − 1)(N − 2) . . . (2)(1)n! = n(n − 1)(n − 2) . . . (2)(1)0! = 1

Number of Permutations of N Objects Taken n at a Time

where: N! = N(N − 1)(N − 2) . . . (2)(1)n! = n(n − 1)(n − 2) . . . (2)(1)0! = 1

P nNn

NN nn

N =⎛⎝⎜

⎞⎠⎟

=−

! !( )!

P nNn

NN nn

N =⎛⎝⎜

⎞⎠⎟

=−

! !( )!

Counting Rule for Permutations

A third useful counting rule enables us to count thenumber of experimental outcomes when n objects are tobe selected from a set of N objects, where the order ofselection is important.

Assigning Probabilities

Classical Method

Relative Frequency Method

Subjective Method

Assigning probabilities based on the assumptionof equally likely outcomes

Assigning probabilities based on experimentationor historical data

Assigning probabilities based on judgment

Classical Method

If an experiment has n possible outcomes, this method would assign a probability of 1/n to each outcome.

Experiment: Rolling a dieSample Space: S = {1, 2, 3, 4, 5, 6}Probabilities: Each sample point has a

1/6 chance of occurring

Example

Relative Frequency Method

Number ofPolishers Rented

Numberof Days

01234

46

18102

Lucas Tool Rental would like to assignprobabilities to the number of car polishersit rents each day. Office records show the followingfrequencies of daily rentals for the last 40 days.

Example: Lucas Tool Rental

Each probability assignment is given bydividing the frequency (number of days) bythe total frequency (total number of days).

Relative Frequency Method

4/40

ProbabilityNumber of

Polishers RentedNumberof Days

01234

46

18102

40

.10

.15

.45

.25

.051.00

Subjective Method

Applying the subjective method, an analyst made the following probability assignments.

Exper. Outcome Net Gain or Loss Probability(10, 8)(10, −2)(5, 8)(5, −2)(0, 8)(0, −2)(−20, 8)(−20, −2)

$18,000 Gain$8,000 Gain

$13,000 Gain$3,000 Gain$8,000 Gain$2,000 Loss

$12,000 Loss$22,000 Loss

.20

.08

.16

.26

.10

.12

.02

.06

An event is a collection of sample points.

The probability of any event is equal to the sum ofthe probabilities of the sample points in the event.

If we can identify all the sample points of anexperiment and assign a probability to each, wecan compute the probability of an event.

Events and Their Probabilities

Events and Their Probabilities

Event M = Markley Oil ProfitableM = {(10, 8), (10, −2), (5, 8), (5, −2)}

P(M) = P(10, 8) + P(10, −2) + P(5, 8) + P(5, −2)= .20 + .08 + .16 + .26= .70

Events and Their Probabilities

Event C = Collins Mining ProfitableC = {(10, 8), (5, 8), (0, 8), (−20, 8)}

P(C) = P(10, 8) + P(5, 8) + P(0, 8) + P(−20, 8)= .20 + .16 + .10 + .02= .48

Some Basic Relationships of Probability

There are some basic probability relationships thatcan be used to compute the probability of an eventwithout knowledge of all the sample point probabilities.

Complement of an Event

Intersection of Two Events

Mutually Exclusive Events

Union of Two Events

The complement of A is denoted by Ac.

The complement of event A is defined to be the eventconsisting of all sample points that are not in A.

Complement of an Event

Event A AcSampleSpace SSampleSpace S

VennDiagram

The union of events A and B is denoted by A ∪ B.

The union of events A and B is the event containingall sample points that are in A or B or both.

Union of Two Events

SampleSpace SSampleSpace SEvent A Event B

Union of Two Events

Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable

M ∪ C = Markley Oil Profitable or Collins Mining Profitable

M ∪ C = {(10, 8), (10, −2), (5, 8), (5, −2), (0, 8), (−20, 8)}P(M ∪ C) = P(10, 8) + P(10, −2) + P(5, 8) + P(5, −2)

+ P(0, 8) + P(−20, 8)= .20 + .08 + .16 + .26 + .10 + .02= .82

The intersection of events A and B is denoted by A ∩ Β.

The intersection of events A and B is the set of allsample points that are in both A and B.

SampleSpace SSampleSpace SEvent A Event B

Intersection of Two Events

Intersection of A and BIntersection of A and B

Intersection of Two Events

Event M = Markley Oil ProfitableEvent C = Collins Mining ProfitableM ∩ C = Markley Oil Profitable

and Collins Mining ProfitableM ∩ C = {(10, 8), (5, 8)}

P(M ∩ C) = P(10, 8) + P(5, 8)

= .20 + .16

= .36

The addition law provides a way to compute theprobability of event A, or B, or both A and B occurring.

Addition Law

The law is written as:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable

M ∪ C = Markley Oil Profitable or Collins Mining Profitable

We know: P(M) = .70, P(C) = .48, P(M ∩ C) = .36Thus: P(M ∪ C) = P(M) + P(C) − P(M ∩ C)

= .70 + .48 − .36= .82

Addition Law

(This result is the same as that obtained earlierusing the definition of the probability of an event.)

Mutually Exclusive Events

Two events are said to be mutually exclusive if theevents have no sample points in common.

Two events are mutually exclusive if, when one eventoccurs, the other cannot occur.

SampleSpace SSampleSpace SEvent A Event B

Mutually Exclusive Events

If events A and B are mutually exclusive, P(A ∩ B) = 0.

The addition law for mutually exclusive events is:

P(A ∪ B) = P(A) + P(B)

there’s no need toinclude “− P(A ∩ B)”

The probability of an event given that another eventhas occurred is called a conditional probability.

A conditional probability is computed as follows :

The conditional probability of A given B is denotedby P(A|B).

Conditional Probability

( )( | )( )

P A BP A BP B

∩=

( )( | )( )

P A BP A BP B

∩=

Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable

We know: P(M ∩ C) = .36, P(M) = .70

Thus:

Conditional Probability

( ) .36( | ) .5143( ) .70

P C MP C MP M

∩= = =

( ) .36( | ) .5143( ) .70

P C MP C MP M

∩= = =

= Collins Mining Profitablegiven Markley Oil Profitable

( | )P C M( | )P C M

Multiplication Law

The multiplication law provides a way to compute theprobability of the intersection of two events.

The law is written as:

P(A ∩ B) = P(B)P(A|B)

Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable

We know: P(M) = .70, P(C|M) = .5143

Multiplication Law

M ∩ C = Markley Oil Profitableand Collins Mining Profitable

Thus: P(M ∩ C) = P(M)P(M|C)= (.70)(.5143)= .36

(This result is the same as that obtained earlierusing the definition of the probability of an event.)

Independent Events

If the probability of event A is not changed by theexistence of event B, we would say that events Aand B are independent.

Two events A and B are independent if:

P(A|B) = P(A) P(B|A) = P(B)or

The multiplication law also can be used as a test to seeif two events are independent.

The law is written as:

P(A ∩ B) = P(A)P(B)

Multiplication Lawfor Independent Events

Multiplication Lawfor Independent Events

Event M = Markley Oil ProfitableEvent C = Collins Mining Profitable

We know: P(M ∩ C) = .36, P(M) = .70, P(C) = .48But: P(M)P(C) = (.70)(.48) = .34, not .36

Are events M and C independent?Does P(M ∩ C) = P(M)P(C) ?

Hence: M and C are not independent.