Probability and Statistics

1. Define probability. 2. Describe the classical, the empirical, and the subjective

approaches to probability. 3. Understand the terms: experiment, event, outcome,

permutations, and combinations.4. Define the terms: conditional probability and joint

probability.5. Calculate probabilities applying the rules of addition and

the rules of multiplication. 6. Use a tree diagram to organize and compute

probabilities.7. Calculate a probability using Bayes’ theorem.8. Determine the number of permutations and the number

of combinations.

Chapter Five: A Survey of Probability Concepts

GOALSWhen you have completed this chapter, you will be able to:

Important: This file is a modified electronic version of Statistical Techniques in Business and Economics © The McGraw-Hill Companies, Inc., 1999

DefinitionsProbability: A measure of the likelihood

that an event in the future will happen; it can only assume a value between 0 and 1, inclusive.

Experiment: The observation of some activity or the act of taking some measurement.

Outcome: A particular result of an experiment.

Event: A collection of one or more outcomes of an experiment.

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Approaches to ProbabilityClassical probability is based on

the assumption that the outcomes of an experiment are equally likely.

Using this classical viewpoint,

Probability of an event = Number of favorable outcomes

Total number of possible outcomes

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P(A) = n(A) n(S)

EXAMPLE 1Consider the experiment of

tossing two coins once.The sample space S = {HH, HT,

TH, TT}Consider the event of one head.Probability of one head = 2/4 =

1/2.

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How many possible outcomes are there in throwing 2 dice?

Mutually Exclusive Events

Mutually Exclusive Events: The occurrence of any one event means that none of the others can occur at the same time.

In EXAMPLE 1, the four possible outcomes are mutually exclusive.

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Collectively Exhaustive EventsCollectively exhaustive: At least

one of the events must occur when an experiment is conducted.

In EXAMPLE 1, the four possible outcomes are collectively exhaustive. In other words, the sum of probabilities = 1 (0.25 + 0.25 + 0.25 + 0.25).

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Relative Frequency ConceptThe probability of an event

happening in the long run is determined by observing what fraction of the time like events happened in the past:

Probability of event = Number of times event occured in the past

Total number of observations

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EXAMPLE 2Throughout her career Professor

Jones has awarded 186 A’s out of the 1200 students she has taught. What is the probability that a student in her section this semester will receive an A?

By applying the relative frequency concept, the probability of an A= 186/1200=0.155

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Subjective ProbabilitySubjective probability: The likelihood

(probability) of a particular event happening that is assigned by an individual based on whatever information is available.

Examples of subjective probability are estimating the probability the Washington Redskins will win the Super Bowl next year and estimating the probability of an earthquake in Los Angeles this year.

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Basic Rules of ProbabilityIf events are mutually exclusive, then

the occurrence of any one of the events precludes any of the other events from occurring.

Rules of addition: If two events A and B are mutually exclusive, the special rule of addition states that the probability of A or B occurring equals the sum of their respective probabilities: P(A or B) = P(A) + P(B)

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EXAMPLE 3Cebu Pacific Air recently supplied the

following information on their commuter flights from Davao to Manila:

Arrival FrequencyEarly 100

On Time 800Late 75

Canceled 25Total 1000

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If A is the event that a flight arrives early, then P(A) = 100/1000 = 0.1

If B is the event that a flight arrives late, then P(B) = 75/1000 = .075.

The probability that a flight is either early or late is P(A or B) = P(A) + P(B) = 0.1 + 0.075 =0.175.

The Complement RuleThe complement rule is used to

determine the probability of an event occurring by subtracting the probability of the event not occurring from 1. If P(A) is the probability of event A and P(~A) is the complement of A, P(A) + P(~A) = 1

OR P(A) = 1- P(~A)

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The Complement Rule continued

A Venn diagram illustrating the complement rule would appear as:

A ~A

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EXAMPLE 4Recall EXAMPLE 3.If C is the event that a flight

arrives on time, then P(C) = 800/1000 = 0.8

If D is the event that a flight is canceled, then P(D) = 25/1000 = 0.025

Use the complement rule to show that the probability of an early (A) or a late (B) flight is 0.175

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EXAMPLE 4 continued

P(A or B) = 1 - P(C or D) = 1 -[0.8 +0.025] =0.175

C.8

D.025

~(C or D) = (A or B) 0.175

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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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The General Rule of AdditionThe Venn Diagram illustrates this

rule:

A and B

A

B

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EXAMPLE 5

In a sample of 500 students, 320 said they had a stereo, 175 said they had a TV, and 100 said they had both:

Stereo 220

Both 100

TV75

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EXAMPLE 5 continued

If a student is selected at random, what is the probability that the student has only a stereo, only a TV, and both a stereo and TV?

P(S) = 220/500 = 0.44P(T) = 75/500 = 0.15P(S and T) = 100/500 = 0.20

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EXAMPLE 5 continued

If a student is selected at random, what is the probability that the student has either a stereo or a TV in his or her room?P(S or T) = P(S) + P(T) - P(S and T) = 0.44 +0.15 -0.20 = 0.39

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Joint ProbabilityJoint Probability is a probability

that measures the likelihood that two or more events will happen concurrently. An example would be the event that a student has both a stereo and TV in his or her dorm room.

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Special Rule of MultiplicationThe 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.

The special rule is written: P(A and B) = P(A)*P(B).

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EXAMPLE 6Chris owns two stocks which are

independent of each other. The probability that stock A increases in value next year is 0.5. The probability that stock B will increase in value next year is 0.7.

What is the probability that both stocks increase in value next year?

P(A and B) = (0.5)(0.7) = 0.35

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EXAMPLE 6 continued

What is the probability that at least one of these stocks increase in value during the next year (this implies that either one can increase or both)?

Thus, P(at least one) = (0.5)(0.3) +

(0.5)(0.7) + (0.7)(0.5)

= 0.85.

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Conditional ProbabilityConditional probability is the

probability of a particular event occurring, given that another event has occurred.

Note: The probability of the event A given that the event B has occurred is denoted by P(A|B).

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General Multiplication Rule

The general rule of multiplication is used to find the joint probability that two events will occur, as it states: 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 B given that A has occurred.

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General Multiplication Rule

The joint probability, P(A and B) is given by the following formula: P(A and B) = P(A)*P(B|A) OR P(A and B) = P(B)*P(A|B)

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EXAMPLE 7The Dean of the College of

Engineering collected the following information about undergraduate students in his college:MAJOR Male Female Total

Electronics 170 110 280

Electrical 120 100 220

Civil 160 70 230

Mechanical 150 120 270

Total 600 400 1000

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EXAMPLE 7 continued

If a student is selected at random, what is the probability that the student is a female electronics major? P(A and F) = 110/1000.

Given that the student is a female, what is the probability that she is an electronics major? P(A|F) = [P(A and F)]/[P(F)] = [110/1000]/[400/1000]

= 0.275

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Tree DiagramsA tree diagram is very useful for

portraying conditional and joint probabilities and is particularly useful for analyzing business decisions involving several stages.

EXAMPLE 8: In a bag containing 7 red chips and 5 blue chips you select 2 chips one after the other without replacement. Construct a tree diagram for this information.

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EXAMPLE 8 continued

R1

B1

R2

B2

R2

B2

7/12

5/12

6/11

5/117/11

4/11

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Bayes’ TheoremBayes’ Theorem is given by the

formula:

P A BP A P B A

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

( ) * ( | )( ) * ( | ) ( ) * ( | )

11 1

1 1 2 2

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EXAMPLE 9Duff Beer Company has received

several complaints that their bottles are under-filled. A complaint was received today but the production manager is unable to identify which of the two Springfield plants (A or B) filled this bottle. What is the probability that the under filled bottle came from plant A?

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EXAMPLE 9 continued

P(A|U) = [(.55)(.03)]/[(.55)(.03)+(.45)(.04)]

= 0.4783.

% of Total Production

% of under-filled bottles

A 55 3

B 45 4

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P A BP A P B A

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

( ) * ( | )( ) * ( | ) ( ) * ( | )

11 1

1 1 2 2

Some Principles of CountingThe Multiplication Formula: If

there are m ways of doing one thing and n ways of doing another thing, there are m x n ways of doing both.

Example 10: Dr. Delong has 10 shirts and 8 ties. How many shirt/tie outfits does he have? (10)(8) = 80.

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Some Principles of CountingPermutation: Any arrangement of

r objects selected from n possible objects.

Note: The order of arrangement is important in permutations.

nnn r

P!

( )r

!

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Some Principles of CountingCombination: The number of

ways to choose r objects from a group of n objects without regard to order.

n rCn

r n r

!

!( ) !

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EXAMPLE 11Coach Thompson must pick five

players among the twelve on the team to comprise the starting lineup. How many different groups are possible? 12C5 = (12!)/[5!(12-5)!] =792

Suppose Coach Thompson must rank them: 12P5 = (12!)/(12-5)! = 95,040.

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Assignment: Write solution in short bond paper.1. Given 2 dice. What is the probability of getting 7 in a throw of two dice?2. What is the probability of getting 7 or an 11in a throw of two dice? 3. A bag contains 5 red and 3 white balls. What is the probability of getting 3 red balls?4. In a toss of a die, what is the probability of getting less than 4 and greater than 2?5. How many permutations are there of the letters in the word “vivid”?6. How many 3-member committess can be formed from 10 persons?

Given 2 dice. What is the probability of getting 7 in a throw of two dice?

A = {(3,4), (4,3), (5,2), (2,5), (1,6), (6,1)}

Then P(A) = 6/36 = 1/6Given 2 dice. What is the probability of getting 7 or 11 in a throw of two dice?

A1 = {(3,4), (4,3), (5,2), (2,5), (1,6), (6,1)}A2 = {(5,6), (6,5)

Then P(A) = P(A1)+P(A2)=6/36 +2/36= 8/36 = 2/9

A bag contains 5 red and 3 white balls. What is the probability of getting 3 red balls?

P(A) = 5C3 =5!/3!2! 8C3 8!/3!5

Then P(A) = 5/28In a toss of a die, what is the probability of getting less than 4 and greater than 2?

A1 = {x/x<4}A2 = {x/x>2}

Then P(A1 and A2) = 1/6

n rCn

r n r

!

!( )

How many permutations are there of the letters in the word “vivid”?

Since there are 2v’s, 2i’s and 1 d, thenP = 5! = 5 ∙4 ∙3 ∙2 ∙1 = 30 2!2!1! 2 ∙1 ∙2 ∙1 ∙1

How many 3-member committess can be formed from 10 persons?

Any three persons regardless of order in which their names appear, hence, this problem is one That involves combination.Thus, = 10! = 10∙9 ∙8 = 120

3!7! 3 ∙2 ∙1n rC

nr n r

!

!( )