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Understanding Probability: Chapter 5 Review and Exercises

This lesson reviews Chapter 5 on probability concepts, focusing on key definitions and vocabulary. Key objectives include summarizing the chapter, completing review exercises, and applying probability rules to various problems involving moviegoers and automobile defects. Explore real-life examples such as gender preferences for films and the chances of defects in vehicles. Engage with essential exercises to enhance your understanding of probability, mutual exclusivity, and conditional probabilities. Homework assignments reinforce learning from pages 308-311.

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Understanding Probability: Chapter 5 Review and Exercises

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  1. Lesson 5 - R Chapter 5 Review

  2. Objectives • Summarize the chapter • Define the vocabulary used • Complete all objectives • Successfully answer any of the review exercises • Use Probability Rules to determine probability of Events

  3. Vocabulary • None new

  4. Example 1 At a recent movie, 1000 patrons (560 females and 440 males) were asked whether or not they liked the film. In was determined that 355 females liked the film and 250 males said they enjoyed it also. If a person is randomly selected from the moviegoers what is the probability that the moviegoers was: a) male? b) a female and liked the film c) a male or someone who disliked the film? d) a male and disliked the film? e) a male given they liked the film? f) someone who liked the film given they were a female? g) Are sex and film preference independent? P(m) = 440/1000 = 0.44 P(F & L) = 0.56 * 355/560 = 0.44 P(m or DL) = 0.44 P(m & DL) = 0.44 P(m|L) = 250/605 = 0. P(L|F) = 355/560 = 0. P(M|L) = P(M)

  5. Example 2 If P(A) = .3 and P(B) = .45 and events A and B are mutually exclusive find P(A or B). Mutually Exclusive  P(A and B) = 0!! P(A or B) = P(A) + P(B) – P(A and B) P(A or B) = 0.3 + 0.45 = 0.75

  6. Example 3 When spot-checked for safety, automobiles are found to defective tires 15% of the time, defective lights 25% of the time, and both defective tires and lights 8% of the time. Find the probability that a randomly chosen car has defective lights given that its tires are found to be defective. P(DL | DT) = P(DL and DT) / P(DT) P(DL | DT) = 0.08 / 0.15 = 0.5333

  7. Summary and Homework • Summary • Homework:pg 308 - 311: 3, 6, 7, 13, 14, 20, 39

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