1 weather forecast psychology games sports chapter 3 elementary statistics larson farber probability...

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Weather forecast

Psychology

Games Sports

Chapter

3

Elementary Statistics

Larson Farber

Probability

Business

Medicine

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{ 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:

The result of a single trial

Important Terms

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Probability 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: The result of a single trial

Choose a car from production line

Another Experiment

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Classical (equally probable outcomes)

Number of outcomes in event EP(E)

Number of outcomes in sample space

Frequency Total

Eevent of Frequency P(E)

Probability blood pressure will decrease after medication

Probability the line will be busy

Empirical

Subjective

Types of Probability

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Two dice are rolled.Describe the sample space.

1st roll

36 outcomes

2nd roll

Tree DiagramsStart

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

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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

Sample Spaces and Probabilities

Two dice are rolled and the sum is noted.

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Complementary Events

The 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.

E E

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

P(E´) = 1 - P(E)

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

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

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

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

Conditional Probability

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Two dice are rolled, find the probability the 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


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Two events A and B are independent if the probability of the occurrence of event B is not affected by the occurrence (or non-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 Events

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

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

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

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

Two dice are rolled. A= first is a 4 and B = second is a 4 P(B)= 1/6 and P(B|A) = 1/6. The events are independent.

Independent Events


Probability

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

1. P(Yes)

2. P(Seattle)

3. P(Miami)

4. P(No, given Miami)

5. P(Not Seattle)

6. P(Seattle, given yes)

7. P(Yes, given Seattle)

8. P(Miami, given Omaha)

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

Contingency Table

One of the responses is selected at random. Find:

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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 TotalYesNoUndTotal 300 450 250

400

1000

= 400 / 1000 = 0.4

= 95 / 250 = 0.38

=250 / 1000 = 0.25

Solutions

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

= 450 / 1000 = 0.45

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100 150 150

125 130 95 350

75 170 5 250

Omaha Seattle Miami Total

Yes

No

Undecided

Total 300 450 250

400

1000

= 1 - 450 / 1000 = 0.55

= 0 / 250 = 0

Solutions

Answers: 5) 0.55 6) 0.375 7) 0.333 8) 0

5. P(Not Seattle)

6. P(Seattle, given yes)

7. P(Yes, given Seattle)

8. P(Miami, given Omaha)

= 150 / 400 = 0.375

=150 / 450 = 0.333

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Are events A= Seattle and B= Yes independent events?

Are events A = Miami and B = Omaha independent events?

Omaha Seattle Miami Total

Yes 100 150 150 400

No 125 130 95 350

Undecided 75 170 5 250

Total 300 450 250 1000

Solutions

P(Yes |Seattle) = 150/450 = 0.333 P(Yes) = 0.4

If events are independent P(B|A) = P(B)

Since 0.333 0.4 the events are NOT independent.

P(Omaha|Miami) = 0 P(Omaha) = 0.3

Since 0 0.3 the events are NOT independent.

If events are independent P(B|A) = P(B)

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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) × P(B|A)

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

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

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

P(A and B) = 5/12 × 4/11 = 5/33= 0.1515

Multiplication Rule

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

A= first die is a 4 and B= second die is a 4.P(A) = 1/6 P(B|A) = 1/6

P(A and B) = 1/6 × 1/6 = 1/36 = 0.028

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

Note for independent events P(B) and P(B|A) are the same.

Multiplication Rule

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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 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.

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Non-Mutually Exclusive Events

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

A = A person is under 25B = A person is a lawyer

A = A person was born in PhiladelphiaB = A person watches 20/20 on TV.

A BNon-mutually exclusiveP(A and B) 0

A and B

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The Addition Rule

The probability that one or the other of two events will occur is: P(A) + P(B) - P(A and B)

A card is drawn from a 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) = 4/52 and P(B) = 26/52 but P( A and B) = 2/52

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

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The Addition Rule

A card is drawn from a deck. Find the probability the card is a king or a 10.A = the card is a king and 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 = 0.054

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

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

Contingency Table

4. P(Miami or Yes)

5. P(Omaha or No)

6. P(Miami or Seattle)

1. P(Miami and Yes)

2. P(Omaha and No)

3. P(Miami and Seattle)



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Contingency Table

1. P(Miami and Yes)

2. P(Omaha and No)

3. P(Miami and Seattle)

= 250/1000 * 150/250 = 150/1000 = 0.15

= 300/1000 * 125/300 = 125/1000 = 0.125

= 0



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Contingency Table

4. P(Miami or Yes)

5. P(Omaha or No)

6. P(Miami or Seattle) 250/1000 + 450/1000 - 0/1000=700/1000 = 0.7

Answers: 4) 0.5 5) 0.525 6) 0.7


250/1000 + 400/1000 - 150/1000=500/1000 = 0.5

300/1000 + 350/1000 - 125/1000=525/1000 = 0.525

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Summary

Probability at least one of two events occur

P(A or B) = P(A) + P(B) - P(A and B)Add the simple probabilities but to prevent double counting, don’t forget to subtract the probability of both occurring

For complementary events P(E') = 1 - P(E)Subtract the probability of the event from one.

The probability both of two events occurP(A and B) = P(A) *P(B|A)

Multiply the probability of the first event by the conditional probability the second event occurs, given the first occurred.

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Fundamental Counting Principle

If one event can occur m ways and a second event can occur n ways, the 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. 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

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PermutationsA 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

)!(

!

rn

nPrn

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

You have 6720 permutations of 8 books reading 5.

6720123

12345678

)!58(

!858

P

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CombinationsA combination is an selection or r objects from a group of n objects.

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

!)!(

!

rrn

nCrn

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.

5612345123

12345678

!5)!58(

!858

C

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Permutations of 4 objects taking 2

Each of the 12 groups represents a permutation.

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Combinations of 4 objects taking 2

Each of the 6 groups represents a combination

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Distinguishable Permutations

• What if when we are selecting items, and order matters and some are the same?

n= the total number of items

n = the number of each choice

Ex- How many ways can we order the letters of illinois?

!!!!

!

321 knnnn

n

k

3360!1!1!1!2!3

!8

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