D3.a How Do I Find The Probability of Simple Independent Events?

Course 3

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Warm UpWrite each fraction in simplest form.

1. 2.

3. 4.

Course 3

D3.aHow Do I Find The Probability of

Simple Independent Events?

1620

1236

864

39195

4

5

1

3

1

8

1

5

Problem of the Day

A careless reader mixed up some encyclopedia volumes on a library shelf. The Q volume is to the right of the X volume, and the C is between the X and D volumes. The Q is to the left of the G. X is to the right of C. From right to left, in what order are the volumes?D, C, X, Q, G

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10-1Probability

How do I find the probability of simple independent events?

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10-1Probability

Vocabularyexperimenttrialoutcomesample spaceeventprobabilityimpossiblecertain

Insert Lesson Title Here

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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

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340 1

Give the probability for each outcome.

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%.

Give the probability for each outcome.

Check It Out: Example 1A

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10-1Probability

The polo team has a 45% chance of winning.

The probability of winning is P(win) = 45% = 0.45. The probabilities must add to 1, so the probability of not winning is P(lose) = 1 – 0.45 = 0.55, or 55%.

Give the probability for each outcome.

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

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

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

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Outcome 1 2 3 4 5 6

Probability

One of the six sides of a cube is labeled 2, so a reasonable estimate of the probability that the spinner will land on 2 is P(2) = . 1

6

Check It Out: Example 1B Continued

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10-1Probability

One of the six sides of a cube is labeled 3, so a reasonable estimate of the probability that the spinner will land on 3 is P(3) = . 1

6

One 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

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One of the six sides of a cube is labeled 5, so a reasonable estimate of the probability that the spinner will land on 5 is P(5) = . 1

6

Check It Out: Example 1B Continued

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10-1Probability

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

6

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

16

16

16

++ = 116

+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.

A quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score.

Example 2A: Finding Probabilities of Events

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10-1Probability

What is the probability of guessing 3 or more correct?

The event “three or more correct” consists of the outcomes 3, 4, and 5.

P(3 or more correct) = 0.313 + 0.156 + 0.031 = 0.5, or 50%.

A quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score.

Check It Out: Example 2A

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10-1Probability

What is the probability of guessing 2 or more correct?

The event “two or more correct” consists of the outcomes 2, 3, 4, and 5.

P(2 or more) = 0.313 + 0.313 + 0.156 + 0.031 = .813, or 81.3%.

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10-1Probability

What is the probability of guessing fewer than 2 correct?

The event “fewer than 2 correct” consists of the outcomes 0 and 1.

P(fewer than 2 correct) = 0.031 + 0.156 = 0.187, or 18.7%

Example 2B: Finding Probabilities of EventsA quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score.

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10-1Probability

What is the probability of guessing fewer than 3 correct?

The event “fewer than 3” consists of the outcomes 0, 1, and 2.

P(fewer than 3) = 0.031 + 0.156 + 0.313 = 0.5, or 50%

Check It Out: Example 2BA quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score.

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10-1Probability

What is the probability of passing the quiz (getting 4 or 5 correct) by guessing?

The event “passing the quiz” consists of the outcomes 4 and 5.

P(passing the quiz) = 0.156 + 0.031 = 0.187, or 18.7%

Example 2C: Finding Probabilities of EventsA quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score.

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10-1Probability

What is the probability of passing the quiz with all 5 correct by guessing?

The event “passing the quiz” consists of the outcome 5.

P(passing the quiz) = 0.031 = or 3.1%

Check It Out: Example 2CA quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score.

Example 3: Problem Solving Application

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10-1Probability

Six students are in a race. Ken’s probability of winning is 0.2. Lee is twice as likely to win as Ken. Roy is as likely to win as Lee. Tracy, James, and Kadeem all have the same chance of winning. Create a table of probabilities for the sample space.

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Example 3 Continued

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10-1Probability

11 Understand the Problem

The answer will be a table of probabilities. Each probability will be a number from 0 to 1. The probabilities of all outcomes add to 1.

List the important information:

• P(Ken) = 0.2

• P(Lee) = 2 P(Ken) = 2 0.2 = 0.4

• P(Tracy) = P(James) = P(Kadeem)

• P(Roy) = P(Lee) = 0.4 = 0.1 14

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Example 3 Continued

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22 Make a Plan

You know the probabilities add to 1, so use the strategy write an equation. Let p represent the probability for Tracy, James, and Kadeem.

P(Ken) + P(Lee) + P(Roy) + P(Tracy) + P(James) + P(Kadeem) = 1

0.2 + 0.4 + 0.1 + p + p + p = 1

0.7 + 3p = 1

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10-1Probability

Solve33

0.7 + 3p = 1

–0.7 –0.7 Subtract 0.7 from both sides.

3p = 0.3

3p3

0.33

= Divide both sides by 3.

Example 3 Continued

p = 0.1

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10-1Probability

Look Back44

Check that the probabilities add to 1.

0.2 + 0.4 + 0.1 + 0.1 + 0.1 + 0.1 = 1

Example 3 Continued

Four students are in the Spelling Bee. Fred’s probability of winning is 0.6. Willa’s chances are one-third of Fred’s. Betty’s and Barrie’s chances are the same. Create a table of probabilities for the sample space.

Check It Out: Example 3

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Check It Out: Example 3 Continued

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10-1Probability

11 Understand the Problem

The answer will be a table of probabilities. Each probability will be a number from 0 to 1. The probabilities of all outcomes add to 1.

List the important information:

• P(Fred) = 0.6

• P(Betty) = P(Barrie)

• P(Willa) = P(Fred) = 0.6 = 0.213

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Check It Out: Example 3 Continued

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10-1Probability

22 Make a Plan

You know the probabilities add to 1, so use the strategy write an equation. Let p represent the probability for Betty and Barrie.

P(Fred) + P(Willa) + P(Betty) + P(Barrie) = 1

0.6 + 0.2 + p + p = 1

0.8 + 2p = 1

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10-1Probability

Solve33

0.8 + 2p = 1

–0.8 –0.8 Subtract 0.8 from both sides.

2p = 0.2

Check It Out: Example 3 Continued

Outcome Fred Willa Betty Barrie

Probability 0.6 0.2 0.1 0.1

p = 0.1

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10-1Probability

Look Back44

Check that the probabilities add to 1.

0.6 + 0.2 + 0.1 + 0.1 = 1

Check It Out: Example 3 Continued

Lesson Quiz

Use the table to find the probability of each event.

1. 1 or 2 occurring

2. 3 not occurring

3. 2, 3, or 4 occurring0.874

0.351

Insert Lesson Title Here

0.794

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10-1Probability