# 10-1 Probability Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation.

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10-1 Probability Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

Warm Up Write each fraction in simplest form. 1.2. 3.4. Course 3 10-1 Probability 16 20 12 36 8 64 39 195 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 Course 3 10-1 Probability

Learn to find the probability of an event by using the definition of probability. Course 3 10-1 Probability

Vocabulary experiment trial outcome sample space event probability impossible certain Insert Lesson Title Here Course 3 10-1 Probability

Course 3 10-1 Probability 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. ExperimentSample 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

Course 3 10-1 Probability 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.

Course 3 10-1 Probability 0 0.25 0.5 0.75 1 0% 25% 50% 75% 100% NeverHappens about Always happenshalf the timehappens 1 4 1 2 3 4 0 1

Give the probability for each outcome. Additional Example 1A: Finding Probabilities of Outcomes in a Sample Space Course 3 10-1 Probability 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. Additional Example 1B: Finding Probabilities of Outcomes in a Sample Space Course 3 10-1 Probability 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

Additional Example 1B Continued Course 3 10-1 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) =. 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 3 8 2 8 ++ = 1

Give the probability for each outcome. Check It Out: Example 1A Course 3 10-1 Probability 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 Course 3 10-1 Probability 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 123456 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 Course 3 10-1 Probability 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 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 5 is P(5) =. 1 6

Check It Out: Example 1B Continued Course 3 10-1 Probability 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. 1 6 1 6 1 6 ++ = 1 1 6 + 1 6 + 1 6 +

Course 3 10-1 Probability 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. Additional Example 2A: Finding Probabilities of Events Course 3 10-1 Probability 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%.

Course 3 10-1 Probability 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% 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.

Course 3 10-1 Probability 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% 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.

Check It Out: Example 2A Course 3 10-1 Probability 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%.

Course 3 10-1 Probability 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 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.

Course 3 10-1 Probability 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 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.

Additional Example 3: Problem Solving Application Course 3 10-1 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. 1414

Additional Example 3 Continued Course 3 10-1 Probability 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(Tracy) = P(James) = P(Kadeem) P(Roy) =  P(Lee) =  0.4 = 0.1 1 4 1 4

Additional Example 3 Continued Course 3 10-1 Probability 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 0.2 + 0.4 + 0.1 + p + p + p = 1 0.7 + 3p = 1

Course 3 10-1 Probability Solve 3 0.7 + 3p = 1 –0.7 –0.7 Subtract 0.7 from both sides. 3p = 0.3 3p3p 3 0.3 3 = Divide both sides by 3. Additional Example 3 Continued p = 0.1

Course 3 10-1 Probability Look Back4 Check that the probabilities add to 1. 0.2 + 0.4 + 0.1 + 0.1 + 0.1 + 0.1 = 1 Additional 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 Course 3 10-1 Probability

Check It Out: Example 3 Continued Course 3 10-1 Probability 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(Betty) = P(Barrie) P(Willa) =  P(Fred) =  0.6 = 0.2 1 3 1 3

Check It Out: Example 3 Continued Course 3 10-1 Probability 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 0.6 + 0.2 + p + p = 1 0.8 + 2p = 1

Course 3 10-1 Probability Solve 3 0.8 + 2p = 1 –0.8 –0.8Subtract 0.8 from both sides. 2p = 0.2 Check It Out: Example 3 Continued OutcomeFredWillaBettyBarrie Probability0.60.20.1 p = 0.1

Course 3 10-1 Probability Look Back4 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 occurring 0.874 0.351 Insert Lesson Title Here 0.794 Course 3 10-1 Probability

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