Probability power point combo from holt ch 10

Various Probability Concepts

Learn to find the probability of an event by using the definition of probability.

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10-1Probability

Vocabularyexperimenttrialoutcomesample spaceeventprobabilityimpossiblecertain

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10-1Probability

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10-1Probability

An experiment is an activity in which results are observed. Each observation is called a trial, and each result is called an outcome. The sample space is the set of all possible outcomes of an experiment.

Experiment Sample Space

flipping a coin heads, tails

rolling a number cube 1, 2, 3, 4, 5, 6

guessing the number of whole numbers marbles in a jar

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10-1Probability

An event is any set of one or more outcomes. The probability of an event, written P(event), is a number from 0 (or 0%) to 1 (or 100%) that tells you how likely the event is to happen.

• A probability of 0 means the event is impossible, or can never happen.

• A probability of 1 means the event is certain, or has to happen.

• The probabilities of all the outcomes in the sample space add up to 1.

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10-1Probability

0 0.25 0.5 0.75 1

0% 25% 50% 75% 100%

Never Happens about Alwayshappens half the time happens

14

12

340 1

Give the probability for each outcome.

Additional Example 1A: Finding Probabilities of Outcomes in a Sample Space

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10-1Probability

The basketball team has a 70% chance of winning.

The probability of winning is P(win) = 70% = 0.7. The probabilities must add to 1, so the probability of not winning is P(lose) = 1 – 0.7 = 0.3, or 30%.


Additional Example 1B: Finding Probabilities of Outcomes in a Sample Space

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10-1Probability

Three of the eight sections of the spinner are labeled 1, so a reasonable estimate of the probability that the spinner will land on 1 is

P(1) = .38

Additional Example 1B Continued

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10-1Probability

Three of the eight sections of the spinner are labeled 2, so a reasonable estimate of the probability that the spinner will land on 2 is P(2) = .3

8

Two of the eight sections of the spinner are labeled 3, so a reasonable estimate of the probability that the spinner will land on 3 is P(3) = = .2

814

Check The probabilities of all the outcomes must add to 1.

38

38

28

++ = 1


Check It Out: Example 1A

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10-1Probability

The polo team has a 50% chance of winning.

The probability of winning is P(win) = 50% = 0.5. The probabilities must add to 1, so the probability of not winning is P(lose) = 1 – 0.5 = 0.5, or 50%.

Give the probability for each outcome.Check It Out: Example 1B

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10-1Probability

Rolling a number cube.

One of the six sides of a cube is labeled 1, so a reasonable estimate of the probability that the spinner will land on 1 is P(1) = . 1

6

Outcome 1 2 3 4 5 6

Probability


6

Check It Out: Example 1B Continued

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10-1Probability


6One of the six sides of a cube is labeled 4, so a reasonable estimate of the probability that the spinner will land on 4 is P(4) = . 1

6


6


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10-1Probability


6

Check The probabilities of all the outcomes must add to 1.

16

16

16++ = 1

16

+ 16

+ 16

+

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10-1Probability

To find the probability of an event, add the probabilities of all the outcomes included in the event.

Learn to find the number of possible outcomes in an experiment.

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10-8 Counting Principles

VocabularyFundamental Counting Principletree diagramAddition Counting Principle


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License plates are being produced that have a single letter followed by three digits. All license plates are equally likely.

Additional Example 1A: Using the Fundamental Counting Principle

Find the number of possible license plates.

Use the Fundamental Counting Principal.

letter first digit second digit third digit

26 choices 10 choices 10 choices 10 choices26 • 10 • 10 • 10 = 26,000

The number of possible 1-letter, 3-digit license plates is 26,000.

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Additional Example 1B: Using the Fundamental Counting Principal

Find the probability that a license plate has the letter Q.

1 • 10 • 10 • 1026,000 = 1

26≈ 0.038P(Q ) =

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Additional Example 1C: Using the Fundamental Counting Principle

Find the probability that a license plate does not contain a 3.

First use the Fundamental Counting Principle to find the number of license plates that do not contain a 3.26 • 9 • 9 • 9 = 18,954 possible license plates without a 3There are 9 choices for any digit except 3.

P(no 3) = = 0.72926,00018,954

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Social Security numbers contain 9 digits. All social security numbers are equally likely.


Find the number of possible Social Security numbers.

Use the Fundamental Counting Principle.

Digit 1 2 3 4 5 6 7 8 9

Choices 10 10 10 10 10 10 10 10 10

10 • 10 • 10 • 10 • 10 • 10 • 10 • 10 • 10 = 1,000,000,000The number of Social Security numbers is 1,000,000,000.

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Check It Out: Example 1B

Find the probability that the Social Security number contains a 7.

P(7 _ _ _ _ _ _ _ _) = 1 • 10 • 10 • 10 • 10 • 10 • 10 • 10 • 10 1,000,000,000

= = 0.1 101

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Check It Out: Example 1C

Find the probability that a Social Security number does not contain a 7.

First use the Fundamental Counting Principle to find the number of Social Security numbers that do not contain a 7.

P(no 7 _ _ _ _ _ _ _ _) = 9 • 9 • 9 • 9 • 9 • 9 • 9 • 9 • 9 1,000,000,000

P(no 7) = ≈ 0.4 1,000,000,000

387,420,489

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The Fundamental Counting Principle tells you only the number of outcomes in some experiments, not what the outcomes are. A tree diagram is a way to show all of the possible outcomes.

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Additional Example 2: Using a Tree Diagram

You have a photo that you want to mat and frame. You can choose from a blue, purple, red, or green mat and a metal or wood frame. Describe all of the ways you could frame this photo with one mat and one frame.

You can find all of the possible outcomes by making a tree diagram.

There should be 4 • 2 = 8 different ways to frame the photo.

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Additional Example 2 Continued

Each “branch” of the tree diagram represents a different way to frame the photo. The ways shown in the branches could be written as (blue, metal), (blue, wood), (purple, metal), (purple, wood), (red, metal), (red, wood), (green, metal), and (green, wood).

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Check It Out: Example 2

A baker can make yellow or white cakes with a choice of chocolate, strawberry, or vanilla icing. Describe all of the possible combinations of cakes.

You can find all of the possible outcomes by making a tree diagram.

There should be 2 • 3 = 6 different cakes available.

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Check It Out: Example 2 Continued

The different cake possibilities are (yellow, chocolate), (yellow, strawberry), (yellow, vanilla), (white, chocolate), (white, strawberry), and (white, vanilla).

white cake

yellow cake

chocolate icing

vanilla icing

strawberry icing

chocolate icing

vanilla icing

strawberry icing

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Additional Example 3: Using the Addition Counting Principle

The table shows the items available at a farm stand. How many items can you choose from the farm stand?

None of the lists contains identical items, so use the Addition Counting Principle.

Total Choices

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Apples Pears Squash+= +

Apples Pears Squash

Macintosh Bosc Acorn

Red Delicious Yellow Bartlett Hubbard

Gold Delicious Red Bartlett

Additional Example 3 Continued

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T 3 3 2+= + = 8

There are 8 items to choose from.

Learn to find the probabilities of independent and dependent events.

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10-5 Independent and Dependent Events

Vocabularycompound events

independent eventsdependent events


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A compound event is made up of one or more separate events. To find the probability of a compound event, you need to know if the events are independent or dependent.

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Events are independent events if the occurrence of one event does not affect the probability of the other. Events are dependent events if the occurrence of one does affect the probability of the other.

Determine if the events are dependent or independent.

A. getting tails on a coin toss and rolling a 6 on a number cube

B. getting 2 red gumballs out of a gumball machine

Additional Example 1: Classifying Events as Independent or Dependent

Tossing a coin does not affect rolling a number cube, so the two events are independent.

After getting one red gumball out of a gumball machine, the chances for getting the second red gumball have changed, so the two events are dependent.

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Determine if the events are dependent or independent.

A. rolling a 6 two times in a row with the same number cube

B. a computer randomly generating two of the same numbers in a row

Check It Out: Example 1

The first roll of the number cube does not affect the second roll, so the events are independent.

The first randomly generated number does not affect the second randomly generated number, so the two events are independent.

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Three separate boxes each have one blue marble and one green marble. One marble is chosen from each box.

What is the probability of choosing a blue marble from each box?

Additional Example 2A: Finding the Probability of Independent Events

The outcome of each choice does not affect the outcome of the other choices, so the choices are independent.

P(blue, blue, blue) =

In each box, P(blue) = .12

12

· 12

· 12

= 18

= 0.125 Multiply.

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What is the probability of choosing at least one blue marble?

Additional Example 2C: Finding the Probability of Independent Events

1 – 0.125 = 0.875

Subtract from 1 to find the probability of choosing at least one blue marble.

Think: P(at least one blue) + P(not blue, not blue, not blue) = 1.

In each box, P(not blue) = .1 2P(not blue, not blue, not blue) =

12

· 12

· 12

= 18

= 0.125 Multiply.

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Two boxes each contain 4 marbles: red, blue, green, and black. One marble is chosen from each box.

What is the probability of choosing a blue marble from each box?


The outcome of each choice does not affect the outcome of the other choices, so the choices are independent.


P(blue, blue) = 14

· 14

= 116

= 0.0625 Multiply.

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What is the probability of choosing a blue marble and then a red marble?



P(blue, red) = 14

· 14

= 116

= 0.0625 Multiply.

In each box, P(red) = .14

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What is the probability of choosing at least one blue marble?

Check It Out: Example 2C


P(not blue, not blue) = 34

· 34

= 916

= 0.5625 Multiply.

Think: P(at least one blue) + P(not blue, not blue) = 1.

1 – 0.5625 = 0.4375Subtract from 1 to find the probability of choosing at least one blue marble.

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To calculate the probability of two dependent events occurring, do the following:

1. Calculate the probability of the first event.

2. Calculate the probability that the second event would occur if the first event had already occurred.

3. Multiply the probabilities.

The letters in the word dependent are placed in a box.

If two letters are chosen at random, what is the probability that they will both be consonants?

Additional Example 3A: Find the Probability of Dependent Events

P(first consonant) =

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23

69

=

Because the first letter is not replaced, the sample space is different for the second letter, so the events are dependent. Find the probability that the first letter chosen is a consonant.

Additional Example 3A Continued

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If the first letter chosen was a consonant, now there would be 5 consonants and a total of 8 letters left in the box. Find the probability that the second letter chosen is a consonant.

P(second consonant) = 58

5 12

58

23

· =

The probability of choosing two letters that are both consonants is .5

12

Multiply.

If two letters are chosen at random, what is the probability that they will both be consonants or both be vowels?

Additional Example 3B: Find the Probability of Dependent Events

There are two possibilities: 2 consonants or 2 vowels. The probability of 2 consonants was calculated in Example 3A. Now find the probability of getting 2 vowels.

Find the probability that the first letter chosen is a vowel.

If the first letter chosen was a vowel, there are now only 2 vowels and 8 total letters left in the box.

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P(first vowel) = 13

39

=

Additional Example 3B Continued

Find the probability that the second letter chosen is a vowel.

The events of both consonants and both vowels are mutually exclusive, so you can add their probabilities.

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P(second vowel) = 14

28

=

1 12

14

13

· = Multiply.

12

5 12

1 12

+ = 6 12

=

The probability of getting two letters that are either both consonants or both vowels is .1

2

P(consonant) + P(vowel)

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Two mutually exclusive events cannot both happen at the same time.

Remember!

The letters in the phrase I Love Math are placed in a box.

If two letters are chosen at random, what is the probability that they will both be consonants?


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P(first consonant) = 59

Because the first letter is not replaced, the sample space is different for the second letter, so the events are dependant. Find the probability that the first letter chosen is a consonant.

Check It Out: Example 3A Continued

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P(second consonant) =

5 18

12

59

· =

The probability of choosing two letters that are both consonants is .5

18

Multiply.

If the first letter chosen was a consonant, now there would be 4 consonants and a total of 8 letters left in the box. Find the probability that the second letter chosen is a consonant.

12

48

=

If two letters are chosen at random, what is the probability that they will both be consonants or both be vowels?


There are two possibilities: 2 consonants or 2 vowels. The probability of 2 consonants was calculated in Try This 3A. Now find the probability of getting 2 vowels.

Find the probability that the first letter chosen is a vowel.

If the first letter chosen was a vowel, there are now only 3 vowels and 8 total letters left in the box.

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P(first vowel) = 49


Find the probability that the second letter chosen is a vowel.

The events of both consonants and both vowels are mutually exclusive, so you can add their probabilities.

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P(second vowel) = 38

12 72

38

49

· = Multiply.16

=

49

5 18

1 6

+ = 8 18

= P(consonant) + P(vowel)

The probability of getting two letters that are either both consonants or both vowels is .4

9

Lesson QuizDetermine if each event is dependent or independent.

1. drawing a red ball from a bucket and then drawing a green ball without replacing the first

2. spinning a 7 on a spinner three times in a row

3. A bucket contains 5 yellow and 7 red balls. If 2 balls are selected randomly without replacement, what is the probability that they will both be yellow?

independent

dependent


533

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Probability power point combo from holt ch 10

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Transcript of Probability power point combo from holt ch 10