# 9-1 Probability Warm Up Problem of the Day Lesson Presentation

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9-1 Probability Warm Up Problem of the Day Lesson Presentation
Pre-Algebra

9-1 Probability Warm Up Write each fraction in simplest form. 1. 2.
Pre-Algebra 9-1 Probability Warm Up Write each fraction in simplest form. 16 20 4 5 12 36 1 3 8 1 8 39 195 1 5 64

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

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

Vocabulary experiment trial outcome sample space event probability
impossible certain

An experiment is an activity in which results are observed
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 jelly beans in a jar

An event is any set of one or more outcomes
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.

happens half the time happens 1 4 1 2 3 4 1 0% % % % %

A. The basketball team has a 70% chance of winning.
Additional Example 1A: Finding Probabilities of Outcomes in a Sample Space Give the probability for each outcome. A. 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
Give the probability for each outcome. B. 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) = . 3 8

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 8 1 4 Check The probabilities of all the outcomes must add to 1. 3 8 2 + = 1

A. The polo team has a 50% chance of winning.
Try This: Example 1A Give the probability for each outcome. A. 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%.

B. Rolling a number cube. 1 2 3 4 5 6 Try This: Example 1B Outcome
Give the probability for each outcome. B. Rolling a number cube. Outcome 1 2 3 4 5 6 Probability 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 One of the six sides of a cube is labeled 2, so a reasonable estimate of the probability that the spinner will land on 1 is P(2) = . 1 6

Try This: Example 1B Continued
One of the six sides of a cube is labeled 3, so a reasonable estimate of the probability that the spinner will land on 1 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 1 is P(4) = . 1 6 One of the six sides of a cube is labeled 5, so a reasonable estimate of the probability that the spinner will land on 1 is P(5) = . 1 6

Try This: Example 1B Continued
One of the six sides of a cube is labeled 6, so a reasonable estimate of the probability that the spinner will land on 1 is P(6) = . 1 6 Check The probabilities of all the outcomes must add to 1. 1 6 + = 1

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

Additional Example 2A: Finding Probabilities of Events
A quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score. A. What is the probability of not guessing 3 or more correct? The event “not three or more correct” consists of the outcomes 0, 1, and 2. P(not 3 or more) = = 0.5, or 50%.

Additional Example 2B: Finding Probabilities of Events
A quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score. B. What is the probability of guessing between 2 and 5? The event “between 2 and 5” consists of the outcomes 3 and 4. P(between 2 and 5) = = 0.469, or 46.9%

Additional Example 2C: Finding Probabilities of Events
A quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score. C. What is the probability of guessing an even number of questions correctly (not counting zero)? The event “even number correct” consists of the outcomes 2 and 4. P(even number correct) = = 0.469, or 46.9%

Try This: Example 2A A quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score. A. 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) = = 0.5, or 50%.

Try This: Example 2B A quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score. B. 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) = = 0.5, or 50%

Try This: Example 2C A quiz contains 5 true or false questions. Suppose you guess randomly on every question. The table below gives the probability of each score. C. 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) = = 0.187, or 18.7%

Additional Example 3: Problem Solving Application
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. 14

Understand the Problem
Additional Example 3 Continued 1 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(Roy) =  P(Lee) =  0.4 = 0.1 1 4 P(Tracy) = P(James) = P(Kadeem)

2 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 p p p = 1 p = 1

Solve 3 p = 1 – –0.7 Subtract 0.7 from both sides. 3p = 0.3 3p 3 0.3 = Divide both sides by 3. p = 0.1

Look Back 4 Check that the probabilities add to 1. = 1

Try This: Example 3 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.

Understand the Problem
Try This: Example 3 Continued 1 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(Willa) =  P(Fred) =  0.6 = 0.2 1 3 P(Betty) = P(Barrie)

Try This: Example 3 Continued
2 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 p p = 1 p = 1

Try This: Example 3 Continued
Solve 3 p = 1 – –0.8 Subtract 0.8 from both sides. 2p = 0.2 p = 0.1 Outcome Fred Willa Betty Barrie Probability 0.6 0.2 0.1

Try This: Example 3 Continued
Look Back 4 Check that the probabilities add to 1. = 1

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 occurring

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