Larson/Farber Ch. 3

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3

Elementary Statistics

Larson Farber

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Probability

Larson/Farber Ch. 3

Section 3.1

Basic Concepts of Probability

Larson/Farber Ch. 3

{ 1 2 3 4 5 6 }

{ Die is even } = { 2 4 6 }

{4}

Roll a dieProbability experiment:An action through which counts, measurements or responses are obtained

Sample space:

The set of all possible outcomes

Event:

A subset of the sample space.

Outcome:

Example

The result of a single trial

Larson/Farber Ch. 3

Simple Event – an event that consists of a single outcome

Example: “tossing heads” and “rolling a 3” is a simple event because there is 1 possible outcome.

“tossing heads and “rolling an even number” is not a simple event because there are 3 possible outcomes

Simple Event

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Classical (theoretical)

Based on educated guesses, intuition and estimates

Empirical (actual, statistical, experimental)

Subjective (intuition)

Types of Probability

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As you increase the number of times a probability experiment is repeated, the empirical probability of the event approaches the theoretical probability of the event.

Law of Large Numbers

Larson/Farber Ch. 3

Two dice are rolled.Describe the sample space.

1st roll

36 outcomes

2nd roll

Start

1 2 3 4 5 6

1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6

Tree Diagrams

Larson/Farber Ch. 3

1,11,21,31,41,51,6

2,12,22,32,42,52,6

3,13,23,33,43,53,6

4,14,24,34,44,54,6

5,15,25,35,45,55,6

6,16,26,36,46,56,6

Find the probability the sum is 4.

Find the probability the sum is 11.

Find the probability the sum is 4 or 11.

Two dice are rolled and the sum is noted.

Sample Space and Probabilities

Larson/Farber Ch. 3

1,11,21,31,41,51,6

2,12,22,32,42,52,6

3,13,23,33,43,53,6

4,14,24,34,44,54,6

5,15,25,35,45,55,6

6,16,26,36,46,56,6

Find the probability the sum is 4.

Find the probability the sum is 11.

Find the probability the sum is 4 or 11.

Two dice are rolled and the sum is noted.

Sample Space and Probabilities

3/36 = 1/12 = 0.083

2/36 = 1/18 = 0.056

5/36 = 0.139

Larson/Farber Ch. 3

Range of Probabilities

The probability of an event E is between 0 and 1, inclusive

Probability cannot be negativeProbability cannot be greater than 1

If probability is 1, the event is certainIf probability is 0, the event is impossible

Larson/Farber Ch. 3

Complementary EventsThe complement of event E is event E´. E´ consists of all the events in the sample space that are not in event E.

The day’s production consists of 12 cars, 5 of which are defective. If one car is selected at random, find the probability it is not defective.

Solution:P(defective) = 5/12P(not defective) = 1 - 5/12 = 7/12 = 0.583

P(E´) = 1 - P(E)

Larson/Farber Ch. 3

Section 3.2

Conditional Probability

and the Multiplication

Rule

Larson/Farber Ch. 3

The probability an event B will occur, given (on the condition) that another event A has occurred.

Two cars are selected from a production line of 12 cars where 5 are defective. What is the probability the 2nd car is defective, given the first car was defective?

We write this as P(B|A) and say “probability of B, given A.”

Conditional Probability

Larson/Farber Ch. 3

Two cars are selected from a production line of 12 cars where 5 are defective. What is the probability the 2nd car is defective, given the first car was defective?

Given a defective car has been selected, the conditional sample space has 4 defective out of 11. P(B|A) = 4/11

Conditional Probability

Larson/Farber Ch. 3

Two events A and B are independent if the probability of the occurrence of event B is not affected by the occurrence of event A.

A = taking an aspirin each day B = having a heart attack

A = being a femaleB = being under 64” tall

Two events that are not independent are dependent.

A = Being femaleB = Having type O blood

A = 1st child is a boyB = 2nd child is a boy

Independent / Dependent Events

Larson/Farber Ch. 3

Two dice are rolled. Find the probabilitythe second die is a 4 given the first was a 4.

Original sample space: {1, 2, 3, 4, 5, 6}

Given the first die was a 4, the conditional

sample space is: {1, 2, 3, 4, 5, 6}

The conditional probability, P(B|A) = 1/6

Independent Events

Larson/Farber Ch. 3

If events A and B are independent, then P(B|A) = P(B)

12 cars are on a production line where 5 are defective and 2 cars are selected at random.

A = first car is defectiveB = second car is defective.

The probability of getting a defective car for the second car depends on whether the first was defective. The events are dependent.

Dependent Events

Conditional Probability

Probability

Larson/Farber Ch. 3

The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is:

1. P(Yes)

2. P(Seattle)

3. P(Miami)

4. P(No, given Miami)

Omaha Seattle Miami TotalYes 100 150 150 400No 125 130 95 350Undecided 75 170 5 250

Total 300 450 250 1000

One of the responses is selected at random. Find:

Contingency Table

Larson/Farber Ch. 3

1. P(Yes)

2. P(Seattle)

3. P(Miami)

4. P(No, given Miami)

100 150 150125 130 95 350 75 170 5 250

Omaha Seattle Miami TotalYesNoUndecidedTotal 300 450 250

400

1000

= 95 / 250 = 0.38

= 250 / 1000 = 0.25

Answers: 1) 0.4 2) 0.45 3) 0.25 4) 0.38

= 450 / 1000 = 0.45

Solutions

= 400 / 1000 = 0.4

Larson/Farber Ch. 3

To find the probability that two events, A and B will occur in sequence, multiply the probability A occurs by the conditional probability B occurs, given A has occurred.

P(A and B) = P(A) x P(B|A)

If events A and B are independent then the rule can be simplified to

Multiplication Rule

P(A and B) = P(A) x P(B)

Larson/Farber Ch. 3

Two cars are selected from a production line of 12 where 5 are defective. Find the probability both cars are defective.

A = first car is defective B = second car is defective.

P(A) = 5/12 P(B|A) = 4/11

P(A and B) = 5/12 x 4/11 = 5/33 = 0.152

Multiplication Rule

Larson/Farber Ch. 3

Two dice are rolled. Find the probability both are 4’s.

Multiplication Rule

Larson/Farber Ch. 3

Two dice are rolled. Find the probability both are 4’s.

P(A and B) = 1/6 x 1/6 = 1/36 = 0.028

When two events A and B are independent, then P (A and B) = P(A) x P(B)

Multiplication Rule

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Section 3.3

The Addition Rule

Larson/Farber Ch. 3

Mutually Exclusive Events

Two events, A and B, are mutually exclusive if they cannot occur in the same trial.

A = A person is under 21 years old B = A person is running for the U.S. Senate

A = A person was born in PhiladelphiaB = A person was born in Houston

A B Mutually exclusiveP(A and B) = 0

When event A occurs it excludes event B in the same trial.

Larson/Farber Ch. 3

Non-Mutually Exclusive Events

If two events can occur in the same trial, they are non-mutually exclusive.

A = A person is under 25 years oldB = A person is a lawyer

A = A person was born in PhiladelphiaB = A person watches Revenge on TV

A BNon-mutually exclusiveP(A and B) ≠ 0

A and B

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Compare “A and B” to “A or B”

The compound event “A and B” means that A and B both occur in the same trial. Use the multiplication rule to find P(A and B).

The compound event “A or B” means either A can occur without B, B can occur without A or both A and B can occur. Use the addition rule to find P(A or B).

A B

A or BA and B

A B

Larson/Farber Ch. 3

The Addition Rule

The probability that one or the other of two events will occur is:

P(A or B) = P(A) + P(B) – P(A and B)

Larson/Farber Ch. 3

The Addition Rule

A card is drawn from a standard deck. Find the probability it is a king or it is red.

A = the card is a king B = the card is red.

P(A or B) = P(A) + P(B) – P(A and B)

P(A) = 4/52, P(B) = 26/52, P(A and B) = 2/52

P(A or B) = 4/52 + 26/52 – 2/52 = 28/52 = 0.538

Larson/Farber Ch. 3

The Addition Rule

A card is drawn from a standard deck. Find the probability the card is a king or a 10.

Larson/Farber Ch. 3

The Addition Rule

A card is drawn from a standard deck. Find the probability the card is a king or a 10.

A = the card is a king B = the card is a 10.

P(A) = 4/52 and P(B) = 4/52 and P(A and B) = 0/52

P(A or B) = 4/52 + 4/52 – 0/52 = 8/52 = 0.154

When events are mutually exclusive, P(A or B) = P(A) + P(B)

Larson/Farber Ch. 3

The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is:

Contingency Table

3. P(Miami or Yes)

4. P(Miami or Seattle)

Omaha Seattle Miami TotalYes 100 150 150 400No 125 130 95 350Undecided 75 170 5 250Total 300 450 250 1000

One of the responses is selected at random. Find:

1. P(Miami and Yes)

2. P(Miami and Seattle)

Larson/Farber Ch. 3

Contingency Table

1. P(Miami and Yes) = 250/1000 • 150/250 = 0.15

2. P(Miami and Seattle) = 0

3. P(Miami or Yes) = 250/1000 + 400/1000 – 150/1000 = 0.5

4. P(Miami or Seattle) = 250/1000 + 450/1000 – 0/1000 = 0.7

Omaha Seattle Miami TotalYes 100 150 150 400No 125 130 95 350Undecided 75 170 5 250Total 300 450 250 1000

One of the responses is selected at random. Find:

Larson/Farber Ch. 3

See page 144 in the text book for a summary of probability

Summary

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Section 3.4

Counting Principles

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If one event can occur m ways and a second event can occur n ways, then number of ways the two events can occur in sequence is m • n.

This rule can be extended for any number of events occurring in a sequence.

Fundamental Counting Principle

Larson/Farber Ch. 3

If a meal consists of 2 choices of soup, 3 main dishes and 2 desserts, how many different meals can be selected?

= 12 meals

Start

2

Soup

• 3

Main

2•

Dessert

Fundamental Counting Principle

Larson/Farber Ch. 3

How many license plates can you make if a license plate consists of

a. 6 letters which can be repeated

b. 6 letters which cannot be repeated

Fundamental Counting Principle

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Factorials

Suppose you want to arrange n objects in order.

There are n choices for 1st place, n – 1 choices for second, then n – 2 choices for third place and so on until there is one choice for last place.

Using the Fundamental Counting Principle, the number of ways of arranging n objects is:

This is called n factorial and written as n!

n(n – 1)(n – 2)…1

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Factorials

The starting lineup for a baseball team consists of nine players. How manydifferent batting orders are possible using the starting lineup?

Larson/Farber Ch. 3

A permutation is an ordered arrangement.

The number of permutations for n objects is n!

n! = n (n – 1) (n – 2)…..3 • 2 • 1

The number of permutations of n objects taken r at a time is:

Permutations

Larson/Farber Ch. 3

You are required to read 5 books from a list of 8. In how many different orders can you do so?

There are 6720 permutations of 8 books reading 5.

Permutations

Larson/Farber Ch. 3

where n1+n2+n3+…….+nk = n

A contractor wants to plant 6 oak trees, 9 maple trees and 5 poplar trees along the street. In how many distinguishable ways can they be planted?

Distinguishable Permutations

Larson/Farber Ch. 3

A combination is a selection of r objects from a group of n objects.

The number of combinations of n objects taken r at a time is:

Combinations

Larson/Farber Ch. 3

You are required to read 5 books from a list of 8. In how many different ways can you choose the books if order does not matter.

There are 56 combinations of 8 objects taking 5.

Combinations

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2 3 Combinations of 4 objects choosing 2

1 4

Each of the 6 groups represents a combination.

1 2

1 3

1 4

3 4

2 4

2 3

Larson/Farber Ch. 3

2 3Permutations of 4 objects choosing 2

1 4

Each of the 12 groups represents a permutation.

1 2 12

1 3 13

2 3 23

1 4

3 4

2 4

14

34

24

Larson/Farber Ch. 3

A word consists of one L, two Es, two Ts, and one R. If the letters are randomly arranged in order, what is the probability that the arrangement spells the word letter?

Applications

Larson/Farber Ch. 3

Find the probability of being dealt five diamonds from a standard deck of playing cards.

Applications