1 probability ernesto a. diaz faculty mathematics department

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ProbabilityProbability

Ernesto A. DiazFaculty

Mathematics Department

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Specific

Deductive vs. Inductive

Inductive Reasoning

General

SpecificConclusion isguaranteed

Deductive Reasoning

GeneralConclusion is probable

Not always can be proved

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Intro to Probabilities

Year - Author Example of work- 1545: Gerolamo Cardano

- 1654: Antoine Gombauld / Pascal / Fermat

- 1662: John Graunt

- 1713: Jacob Bernoulli

- 1812: Marquis de Laplace

- 1865: Gregor Mendel

-Study of Probability and Gambling

- Probability theory

-Observations on the Bills of Death

- “The art of guessing” applications on government, economics, law, genetics- Analytic Theory of Probabilities. Interpreting scientific data- Foundation of Genetics

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Theoretical ProbabilityConcepts

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Definitions An experiment is a process by

which an observation or outcome is obtained

The possible results of an experiment are called its outcomes.

Sample Space is the set S of all possible outcomes

An event is a subset E of the sample space S.

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Example A dice.

E1={ 3 } “A three comes up” E1 ={ 2, 4, 6 } “an even number”

Events and Outcomes are not the same An event is a subset of the sample

space An outcome is an element of the

sample space

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Definitions continued Theoretical probability (a priori)

based on deductive thinking. It is determined through a study of the possible outcome that can occur for the given experiment.

Empirical probability (a posteriori) based on inductive thinking. It is the relative frequency of occurrence of an event and is determined by actual observations of an experiment.

Subjective It is based on individual experience

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Theoretical Probability If each outcome of an experiment

has the same chance of occurring as any other outcome, they are said to be equally likely outcomes.

number of favorable outcomes in event

( )total number of outcomes in the sample space

EP E

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Example A die is rolled. Find the probability

of rolling a) a 3. b) an odd number c) a number less than 4 d) a 8. e) a number less than 9.

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Solutions a)

b) There are three ways an odd number can occur 1, 3 or 5.

c) Three numbers are less than 4.

number of outcomes that will result in a 2 1(2)

total number of possible outcomes 6P

3 1(odd)

6 2P

3 1(number less than 4)

6 2P

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Solutions: continued

d) There are no outcomes that will result in an 8.

e) All outcomes are less than 10. The event must occur and the probability is 1.

0(number greater than 8) 0

6P

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Empirical Probability

Example: In 100 tosses of a fair die, 19 landed showing a 3. Find the empirical probability of the die landing showing a 3.

Let E be the event of the die landing showing a 3.

number of times event has occurred( )

total number of times the experiment has been performed

EP E

19( ) 0.19

100P E

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The Law of Large Numbers

The law of large numbers states that probability statements apply in practice to a large number of trials, not to a single trial. It is the relative frequency over the long run that is accurately predictable, not individual events or precise totals.

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Important Facts The probability of an event that

cannot occur is 0. The probability of an event that must

occur is 1. Every probability is a number

between 0 and 1 inclusive; that is, 0 P(E) 1.

The sum of the probabilities of all possible outcomes of an experiment is 1.

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XIX Century: Opposition & Synthesis

Adolphe Quetelet James Clerk Maxwell

- Statistics in Social Science

- Work: - Patterns in human traits (e.g. height) follow normal curve- Social Statistics have similarities, e.g. rate of murder vs. suicide in Belgium

- Statistics in Mechanics of Gases

- Work: - Patterns in molecule behavior follow statistical trends

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XIX Century: Opposition & Synthesis

Adolphe Quetelet James Clerk Maxwell

- Statistics in Social Science

- Work: - Patterns in human traits (e.g. height) follow normal curve- Social Statistics have similarities, e.g. rate of murder vs. suicide in Belgium

- Conclusions: - Constant social causes dictate behavior; are individuals free?- Laplace: with sufficient knowledge, nothing is uncertain

- Statistics in Mechanics of Gases

- Work: - Patterns in molecule behavior follow statistical trends

- Conclusions: - Statistical regularities in the large scale say nothing of the behavior of individual in the small scale

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Example A standard deck of cards is well

shuffled. Find the probability that the card is selected.

a) a 10. b) not a 10. c) a heart. d) a ace, one or 2. e) diamond and spade. f) a card greater than 4 and less than 7.

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Example continued a) a 10

There are four 10’s in a deck of 52 cards.

b) not a 10

4 1(10)

52 13P

(not a 10) 1 (10)

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1312

13

P P

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Example continued c) a heart

There are 13 hearts in the deck.

d) an ace, 1 or 2

There are 4 aces, 4 ones and 4 twos, or a total of 12 cards.

13 1(heart)

52 4P

12 3(A, 1 or 2)

52 13P

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Example continued d) diamond and spade

The word and means both events must occur. This is not possible.

e) a card greater than 4 and less than 7

The cards greater than 4 and less than 7 are 5’s, and 6’s.

0(diamond & spade) 0

52P 8 2

( )52 13

P E

Copyright © 2005 Pearson Education, Inc.

12.4

Expected Value (Expectation)

Slide 12-22Copyright © 2005 Pearson Education, Inc.

Expected Value

The symbol P1 represents the probability that the first event will occur, and A1 represents the net amount won or lost if the first event occurs.

1 1 2 2 3 3 ... n nE P A P A P A P A

Slide 12-23Copyright © 2005 Pearson Education, Inc.

Example

Teresa is taking a multiple-choice test in which there are four possible answers for each question. The instructor indicated that she will be awarded 3 points for each correct answer and she will lose 1 point for each incorrect answer and no points will be awarded or subtracted for answers left blank. If Teresa does not know the correct answer to a

question, is it to her advantage or disadvantage to guess?

If she can eliminate one of the possible choices, is it to her advantage or disadvantage to guess at the answer?

Slide 12-24Copyright © 2005 Pearson Education, Inc.

Solution

Expected value if Teresa guesses.

1(guesses correctly)

4P

3(guesses incorrectly)

4P

1 3Teresa's expectation = (3) ( 1)

4 43 3

04 4

Slide 12-25Copyright © 2005 Pearson Education, Inc.

Solution continued—eliminate a choice

1(guesses correctly)

3P

2(guesses incorrectly)

3P

1 2Teresa's expectation = (3) ( 1)

3 32 1

13 3

Slide 12-26Copyright © 2005 Pearson Education, Inc.

Example: Winning a Prize

When Calvin Winters attends a tree farm event, he is given a free ticket for the $75 door prize. A total of 150 tickets will be given out. Determine his expectation of winning the door prize.

1 149Expectation = (75) (0)

150 1501

2

Slide 12-27Copyright © 2005 Pearson Education, Inc.

Example

When Calvin Winters attends a tree farm event, he is given the opportunity to purchase a ticket for the $75 door prize. The cost of the ticket is $3, and 150 tickets will be sold. Determine Calvin’s expectation if he purchases one ticket.

Slide 12-28Copyright © 2005 Pearson Education, Inc.

Solution

Calvin’s expectation is $2.49 when he purchases one ticket.

1 149Expectation = (73) ( 3)

150 15073 447

150 150374

1502.49

Slide 12-29Copyright © 2005 Pearson Education, Inc.

Fair Price

Fair price = expected value + cost to play

Slide 12-30Copyright © 2005 Pearson Education, Inc.

Example

Suppose you are playing a game in which you spin the pointer shown in the figure, and you are awarded the amount shown under the pointer. If is costs $10 to play the game, determine

a) the expectation of the person who plays the game.

b) the fair price to play the game.

$10

$10

$2

$2

$20$15

Slide 12-31Copyright © 2005 Pearson Education, Inc.

Solution

$0

3/8

$10

$10$5$8Amt. Won/Lost

1/81/83/8Probability

$20$15$2Amt. Shown on Wheel

3 3 1 1Expectation = ( $8) ($0) ($5) ($10)

8 8 8 824 5 10

08 8 89

1.125 1.138

Slide 12-32Copyright © 2005 Pearson Education, Inc.

Solution

Fair price = expectation + cost to play

= $1.13 + $10

= $8.87

Thus, the fair price is about $8.87.

Copyright © 2005 Pearson Education, Inc.

12.6

Or and And Problems

Slide 12-34Copyright © 2005 Pearson Education, Inc.

Or Problems

P(A or B) = P(A) + P(B) P(A and B) Example: Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9,

and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains an even number or a number greater than 5.

Slide 12-35Copyright © 2005 Pearson Education, Inc.

Solution

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

Thus, the probability of selecting an even number or a number greater than 5 is 7/10.

even or even and (even) (greater 5)

greater than 5 greater than 5

5 5 3

10 10 107

10

P P P

Slide 12-36Copyright © 2005 Pearson Education, Inc.

Example

Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains a number less than 3 or a number greater than 7.

Slide 12-37Copyright © 2005 Pearson Education, Inc.

Solution

There are no numbers that are both less than 3 and greater than 7. Therefore,

2(less than 3)

10P

3(greater than 7)

10P

2 3 5 10

10 10 10 2

Slide 12-38Copyright © 2005 Pearson Education, Inc.

Mutually Exclusive

Two events A and B are mutually exclusive if it is impossible for both events to occur simultaneously.

Slide 12-39Copyright © 2005 Pearson Education, Inc.

Example

One card is selected from a standard deck of playing cards. Determine the probability of the following events. a) selecting a 3 or a jack b) selecting a jack or a heart c) selecting a picture card or a red card d) selecting a red card or a black card

Slide 12-40Copyright © 2005 Pearson Education, Inc.

Solutions

a) 3 or a jack

b) jack or a heart

4 4(3) ( jack)

52 528 2

52 13

P P

jack and 4 13 1( jack) (heart)

heart 52 52 52

16 4

52 13

P P P

Slide 12-41Copyright © 2005 Pearson Education, Inc.

Solutions continued

c) picture card or red card

d) red card or black card

26 26(red) (black)

52 5252

152

P P

picture & 12 26 6(picture) (red)

red card 52 52 52

32 8

52 13

P P P

Slide 12-42Copyright © 2005 Pearson Education, Inc.

And Problems

P(A and B) = P(A) • P(B) Example: Two cards are to be selected with

replacement from a deck of cards. Find the probability that two red cards will be selected.

( ) ( ) ( ) ( )

26 26

52 521 1 1

2 2 4

P A P B P red P red

Slide 12-43Copyright © 2005 Pearson Education, Inc.

Example

Two cards are to be selected without replacement from a deck of cards. Find the probability that two red cards will be selected.

( ) ( ) ( ) ( )

26 25

52 521 25 25

2 52 104

P A P B P red P red

Slide 12-44Copyright © 2005 Pearson Education, Inc.

Independent Events

Event A and Event B are independent events if the occurrence of either event in no way affects the probability of the occurrence of the other event.

Experiments done with replacement will result in independent events, and those done without replacement will result in dependent events.

Slide 12-45Copyright © 2005 Pearson Education, Inc.

Example

A package of 30 tulip bulbs contains 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers. Three bulbs are randomly selected and planted. Find the probability of each of the following. All three bulbs will produce pink flowers. The first bulb selected will produce a red flower, the

second will produce a yellow flower and the third will produce a red flower.

None of the bulbs will produce a yellow flower. At least one will produce yellow flowers.

Slide 12-46Copyright © 2005 Pearson Education, Inc.

Solution

30 tulip bulbs, 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers.

All three bulbs will produce pink flowers.

3 pink (pink 1) (pink 2) (pink 3)

6 5 4=

30 29 281

203

P P P P

Slide 12-47Copyright © 2005 Pearson Education, Inc.

Solution

30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers.

The first bulb selected will produce a red flower, the second will produce a yellow flower and the third will produce a red flower. red,yellow,red (red) (yellow) (red)

14 10 13=

30 29 2813

174

P P P P

Slide 12-48Copyright © 2005 Pearson Education, Inc.

Solution

30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers.

None of the bulbs will produce a yellow flower.

first not second not third notnone yellow

yellow yellow yellow

20 19 18=

30 29 2857

203

P P P P

Slide 12-49Copyright © 2005 Pearson Education, Inc.

Solution

30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers.

At least one will produce yellow flowers.

P(at least one yellow) = 1 P(no yellow)

571

203146

203