A Survey of Probability Concepts

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Definitions

A probability is a measure of the likelihood that an event in the future will happen. It can only assume a value between 0 and 1.

A value near zero means the event is not likely to happen. A value near one means it is likely.

There are three ways of assigning probability:– classical, – empirical, and – subjective.

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Definitions continued

An experiment is a situation involving chance or probability that leads to results called outcomes.

An outcome is the result of a single trial of an experiment.

An event is one or more outcomes of an experiment.

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Assigning Probabilities

Three approaches to assigning probabilities– Classical– Empirical – Subjective

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Classical Probability

Consider an experiment of rolling a six-sided die. What is the probability of the event “an even number of spots appear face up”?

The possible outcomes are:

There are three “favorable” outcomes (a two, a four, and a six) in the collection of six equally likely possible outcomes.

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Mutually Exclusive Events

Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time.

Events are independent if the occurrence of one event does not affect the

occurrence of another.

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Collectively Exhaustive Events

Events are collectively exhaustive if at least one of the events must occur when an experiment is conducted.

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Empirical Probability

The empirical approach to probability is based on what is called the law of large numbers. The key to establishing probabilities empirically is that more observations will provide a more accurate estimate of the probability.

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Law of Large Numbers

Suppose we toss a fair coin. The result of each toss is either a head or a tail. If we toss the coin a great number of times, the probability of the outcome of heads will approach .5. The following table reports the results of an experiment of flipping a fair coin 1, 10, 50, 100, 500, 1,000 and 10,000 times and then computing the relative frequency of heads

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Empirical Probability - Example

On February 1, 2003, the Space Shuttle Columbia exploded. This was the second disaster in 113 space missions for NASA. On the basis of this information, what is the probability that a future mission is successfully completed?

98.0113

111

flights ofnumber Total

flights successful ofNumber flight successful a ofy Probabilit

Problem

Two dice are rolled, find the probability that the sum is

a) equal to 1

b) equal to 4

c) less than 13

Solution

a) The sample space S of two dice is shown below. S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6) (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) } Let E be the event "sum equal to 1". There are no

outcomes which correspond to a sum equal to 1, hence P(E) = n(E) / n(S) = 0 / 36 = 0

b) Three possible ouctcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.

P(E) = n(E) / n(S) = 3 / 36 = 1 / 12 c) All possible ouctcomes, E = S, give a sum less

than 13, hence. P(E) = n(E) / n(S) = 36 / 36 = 1

Problem

A die is rolled and a coin is tossed, find the probability that the die shows an odd number and the coin shows a head.

Solution

The sample space S of the experiment is as follows S = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H) (1,T),(2,T),(3,T),(4,T),(5,T),(6,T)} Let E be the event "the die shows an odd number and the

coin shows a head". Event E may be described as follows E={(1,H),(3,H),(5,H)} The probability P(E) is given by P(E) = n(E) / n(S) = 3 / 12 = 1 / 4

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Rules for Computing Probabilities

Rules of Addition Special Rule of Addition - If two events A

and B are mutually exclusive, the probability of one or the other event’s occurring equals the sum of their probabilities. P(A or B) = P(A) + P(B)

The General Rule of Addition - If A and B are two events that are not mutually exclusive, then P(A or B) is given by the following formula:P(A or B) = P(A) + P(B) - P(A and B)

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Addition Rule - Example

What is the probability that a card chosen at random from a standard deck of cards will be either a king or a heart?

P(A or B) = P(A) + P(B) - P(A and B)

= 4/52 + 13/52 - 1/52

= 16/52, or .3077

Example

If the probabilities are respectively 0.09, 0.15, 0.21 and 0.23 that a person purchasing a new automobile will choose the color green, white, red or blue, what is the probability that a given buyer will purchase a new automobile that comes in one of those colors?

Solution

Let G,W,R and B be the events that a buyer selects respectively, a green, white, red, or blue automobiles. Since these four events are mutually exclusive, the probability is

P(G U W U R U B)=P(G)+P(W)+P(R)+P(B)

=0.09+0.15+0.21+0.23

=0.68

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The Complement Rule

The complement rule is used to determine the probability of an event occurring by subtracting the probability of the event not occurring from 1.

P(A) + P(~A) = 1

or P(A) = 1 - P(~A).

Example

If the probabilities that an automobile mechanic will service 3,4,5,6,7, or 8 or more cars on any given workday are respectively, 0.12, 0.19, 0.28, 0.24, 0.10 and 0.07, what is the probability that he will service at least 5 cars on his next day at work?

Solution

Let E be the event that at least 5 cars are serviced. Now, P(E)=1-P(Ec)

where, Ec is the event that fewer than 5 cars are serviced.

Since

P(Ec)=0.12+0.19=0.31

Hence, P(E)=1-0.31=0.69

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Joint Probability – Venn Diagram

JOINT PROBABILITY A probability that measures the likelihood two or more events will happen concurrently.

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Special Rule of Multiplication

The special rule of multiplication requires that two events A and B are independent.

Two events A and B are independent if the occurrence of one has no effect on the probability of the occurrence of the other.

This rule is written: P(A and B) = P(A)P(B)

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Multiplication Rule-Example

A survey by the American Automobile association (AAA) revealed 60 percent of its members made airline reservations last year. Two members are selected at random. What is the probability both made airline reservations last year?

Solution:The probability the first member made an airline reservation last year is .60,

written as P(R1) = .60The probability that the second member selected made a reservation is also .60, so

P(R2) = .60.Since the number of AAA members is very large, you may assume thatR1 and R2 are independent.

P(R1 and R2) = P(R1)P(R2) = (.60)(.60) = .36

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Conditional Probability

A conditional probability is the probability of a particular event occurring, given that another event has occurred.

The probability of the event A given that the event B has occurred is written P(A|B).

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General Multiplication Rule

It states that for two events, A and B, the joint probability that both events will happen is found by multiplying the probability that event A will happen by the conditional probability of event B occurring given that A has occurred.

• Conditional Probability - Probability of A given B

Independent events:

• Conditional Probability - Probability of A given B

Independent events:

0)( ,)(

)()( BPwhereBP

BAPBAP

P A B P A

P B A P B

( ) ( )

( ) ( )

Conditional Probability

Rules of conditional probability:Rules of conditional probability:

If events A and D are statistically independent:

so

so

P A B P A BP B

( ) ( )( )

P A B P A B P B

P B A P A

( ) ( ) ( )

( ) ( )

P AD P A

P D A P D

( ) ( )

( ) ( )

P A D P A P D( ) ( ) ( )

Conditional Probability (continued)

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General Multiplication Rule - Example

A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white and the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it on. He plays golf two days in a row and does not do laundry.

What is the likelihood both shirts selected are white?

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The event that the first shirt selected is white is W1. The probability is P(W1) = 9/12

The event that the second shirt selected is also white is identified as W2. The conditional probability that the second shirt selected is white, given that the first shirt selected is also white, is P(W2 | W1) = 8/11.

To determine the probability of 2 white shirts being selected we use formula: P(AB) = P(A) P(B|A)

P(W1 and W2) = P(W1)P(W2 |W1) = (9/12)(8/11) = 0.55

General Multiplication Rule - Example

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Contingency Tables

A CONTINGENCY TABLE is a table used to classify sample observations according to two or more identifiable characteristics

E.g. A survey of 150 adults classified each as to gender and the number of movies attended last month. Each respondent is classified according to two criteria—the number of movies attended and gender.

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Contingency Tables - Example

A sample of executives were surveyed about their loyalty to their company. One of the questions was, “If you were given an offer by another company equal to or slightly better than your present position, would you remain with the company or take the other position?” The responses of the 200 executives in the survey were cross-classified with their length of service with the company.

What is the probability of randomly selecting an executive who is loyal to the company (would remain) and who has more than 10 years of service?

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Event A1 happens if a randomly selected executive will remain with the company despite an equal or slightly better offer from another company. Since there are 120 executives out of the 200 in the survey who would remain with the company

P(A1) = 120/200, or .60.Event B4 happens if a randomly selected executive has more than

10 years of service with the company. Thus, P(B4| A1) is the conditional probability that an executive with more than 10 years of service would remain with the company. Of the 120 executives who would remain 75 have more than 10 years of service, so P(B4| A1) = 75/120.

Contingency Tables - Example

AT& T IBM Total

Telecommunication 40 10 50

Computers 20 30 50

Total 60 40 100

Counts

AT& T IBM Total

Telecommunication .40 .10 .50

Computers .20 .30 .50

Total .60 .40 1.00

Probabilities

2.050.0

10.0

)(

)()(

TP

TIBMPTIBMP

Probability that a project is undertaken by IBM given it is a telecommunications project:

Contingency Table - Example

The probability of the union of several independent events is 1 minus the product of probabilities of their complements:

P A A A An P A P A P A P An( ) ( ) ( ) ( ) ( )1 2 3

11 2 3

The probability of the intersection of several independent events is the product of their separate individual probabilities:

P A A A An P A P A P A P An( ) ( ) ( ) ( ) ( )1 2 3 1 2 3

Product Rules for Independent Events

P A P A B P A B( ) ( ) ( )

In terms of conditional probabilities:

More generally (where Bi make up a partition):

P A P A B P A BP A B P B P A B P B

( ) ( ) ( )( ) ( ) ( ) ( )

P A P A Bi

P ABi

P Bi

( ) ( )

( ) ( )

The Law of Total Probability and Bayes’ Theorem

The law of total probability:

Event U: Stock market will go up in the next yearEvent W: Economy will do well in the next year

66.06.60.)20)(.30(.)80)(.75(.

)()()()()()()(

2.8.1)(80.)(

30.)(

75.)(

WPWUPWPWUPWUPWUPUP

WPWP

WUP

WUP

The Law of Total Probability-Example

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Bayes’ Theorem

Bayes’ Theorem is a method for revising a probability given additional information.

It is computed using the following formula:

An economist believes that during periods of high economic growth, the U.S. dollar appreciates with probability 0.70; in periods of moderate economic growth, the dollar appreciates with probability 0.40; and during periods of low economic growth, the dollar appreciates with probability 0.20.

During any period of time, the probability of high economic growth is 0.30, the probability of moderate economic growth is 0.50, and the probability of low economic growth is 0.50.

Suppose the dollar has been appreciating during the present period. What is the probability we are experiencing a period of high economic growth?

Partition:H - High growth P(H) = 0.30M - Moderate growth P(M) = 0.50L - Low growth P(L) = 0.20

Event A Appreciation

P A HP A MP A L

( ) .( ) .( ) .

0 700 40

0 20

Bayes’ Theorem Extended - Example

P H AP H A

P AP H A

P H A P M A P L AP A H P H

P A H P H P A M P M P A L P L

( )( )

( )( )

( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( . )( . )

( . )( . ) ( . )( . ) ( . )( . ).

. . ...

.

0 70 0 300 70 0 30 0 40 050 0 20 0 20

0 210 21 0 20 0 04

0 210 45

0 467

Example 2-11 (continued)

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Bayes Theorem - Example

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Bayes Theorem – Example (cont.)

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Bayes Theorem – Example (cont.)

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Bayes Theorem – Example (cont.)

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Bayes Theorem – Example (cont.)

Example

Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on the day of Marie's wedding?

Solution:

The sample space is defined by two mutually-exclusive events - it rains or it does not rain. Additionally, a third event occurs when the weatherman predicts rain. Notation for these events appears below.

Event A1. It rains on Marie's wedding. Event A2. It does not rain on Marie's wedding Event B. The weatherman predicts rain. In terms of probabilities, we know the following: P( A1 ) = 5/365 =0.0136985 [It rains 5 days out of the year.] P( A2 ) = 360/365 = 0.9863014 [It does not rain 360 days out of the

year.] P( B | A1 ) = 0.9 [When it rains, the weatherman predicts rain 90% of

the time.] P( B | A2 ) = 0.1 [When it does not rain, the weatherman predicts rain

10% of the time.]

We want to know P( A1 | B ), the probability it will rain on the day of Marie's wedding, given a forecast for rain by the weatherman. The answer can be determined from Bayes' theorem, as shown below.

P( A1 | B ) = P( A1 ) P( B | A1 ) --------------------------------------------- P( A1 ) P( B | A1 ) + P( A2 ) P( B | A2 )

P( A1 | B ) = (0.014)(0.9) / [ (0.014)(0.9) + (0.986)(0.1)] P( A1 | B ) = 0.111