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9-1. Probability. Course 3. Warm Up. Problem of the Day. Lesson Presentation. 9-1. Probability. 16. 12. 1. 1. 1. 4. 20. 36. 5. 8. 3. 5. 39. 195. Course 3. Warm Up Write each fraction in simplest form. 1. 2. 3. 4. 8. 64. 9-1. Probability. Course 3.

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**9-1**Probability Course 3 Warm Up Problem of the Day Lesson Presentation**9-1**Probability 16 12 1 1 1 4 20 36 5 8 3 5 39 195 Course 3 Warm Up Write each fraction in simplest form. 1. 2. 3. 4. 8 64**9-1**Probability Course 3 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**9-1**Probability Course 3 Learn to find the probability of an event by using the definition of probability.**9-1**Probability Course 3 Insert Lesson Title Here Vocabulary experiment trial outcome sample space event probability impossible certain**9-1**Probability Course 3 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**9-1**Probability Course 3 • 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.**9-1**Probability 1 1 3 4 2 4 Course 3 Never Happens about Always happens half the time happens 1 0 0 0.25 0.5 0.75 1 0% 25% 50% 75% 100%**9-1**Probability Course 3 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%.**9-1**Probability 3 8 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) = . Course 3 Additional Example 1B: Finding Probabilities of Outcomes in a Sample Space Give the probability for each outcome. B.**9-1**Probability 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) = . 2 3 3 2 1 3 8 8 8 8 8 4 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) = = . + + = 1 Course 3 Additional Example 1B Continued Check The probabilities of all the outcomes must add to 1. **9-1**Probability Course 3 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%.**9-1**Probability 1 1 6 6 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) = . 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) = . Course 3 Try This: Example 1B Give the probability for each outcome. B. Rolling a number cube.**9-1**Probability 1 1 1 6 6 6 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) = . 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) = . 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) = . Course 3 Try This: Example 1B Continued**9-1**Probability 1 1 1 1 1 1 6 6 6 6 6 6 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 + + + + + = 1 6 Course 3 Try This: Example 1B Continued Check The probabilities of all the outcomes must add to 1. **9-1**Probability Course 3 To find the probability of an event, add the probabilities of all the outcomes included in the event.**9-1**Probability Course 3 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) = 0.031 + 0.156 + 0.313 = 0.5, or 50%.**9-1**Probability Course 3 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) = 0.313 + 0.156 = 0.469, or 46.9%**9-1**Probability Course 3 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) = 0.313 + 0.156 = 0.469, or 46.9%**9-1**Probability Course 3 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) = 0.313 + 0.156 + 0.031 = 0.5, or 50%.**9-1**Probability Course 3 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) = 0.031 + 0.156 + 0.313 = 0.5, or 50%**9-1**Probability Course 3 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) = 0.156 + 0.031 = 0.187, or 18.7%**9-1**Probability 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 Course 3 Additional Example 3: Problem Solving Application**9-1**Probability 1 1 1 4 4 • P(Roy) = P(Lee) = 0.4 = 0.1 Understand the Problem Course 3 Additional Example 3 Continued 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)**9-1**Probability Make a Plan 2 Course 3 Additional Example 3 Continued 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**9-1**Probability 3 Solve 3p 0.3 3 3 = Divide both sides by 3. Course 3 Additional Example 3 Continued 0.7 + 3p = 1 –0.7 –0.7Subtract 0.7 from both sides. 3p = 0.3 p = 0.1**9-1**Probability 4 Look Back Course 3 Additional Example 3 Continued Check that the probabilities add to 1. 0.2 + 0.4 + 0.1 + 0.1 + 0.1 + 0.1 = 1 **9-1**Probability Course 3 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.**9-1**Probability 1 1 1 3 3 • P(Willa) = P(Fred) = 0.6 = 0.2 Understand the Problem Course 3 Try This: Example 3 Continued 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)**9-1**Probability Make a Plan 2 Course 3 Try This: Example 3 Continued 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**9-1**Probability 3 Solve Course 3 Try This: Example 3 Continued 0.8 + 2p = 1 –0.8 –0.8Subtract 0.8 from both sides. 2p = 0.2 p = 0.1**9-1**Probability 4 Look Back Course 3 Try This: Example 3 Continued Check that the probabilities add to 1. 0.6 + 0.2 + 0.1 + 0.1 = 1 **9-1**Probability Course 3 Insert Lesson Title Here 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 0.351 0.874 0.794

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