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The Practice of Statistics, 4 th edition – For AP* STARNES, YATES, MOORE

Chapter 5: Probability: What are the Chances?. Section 5.3 Conditional Probability and Independence. The Practice of Statistics, 4 th edition – For AP* STARNES, YATES, MOORE. Chapter 5 Probability: What Are the Chances?. 5.1 Randomness, Probability, and Simulation 5.2 Probability Rules

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The Practice of Statistics, 4 th edition – For AP* STARNES, YATES, MOORE

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  1. Chapter 5: Probability: What are the Chances? Section 5.3 Conditional Probability and Independence The Practice of Statistics, 4th edition – For AP* STARNES, YATES, MOORE

  2. Chapter 5Probability: What Are the Chances? • 5.1 Randomness, Probability, and Simulation • 5.2 Probability Rules • 5.3 Conditional Probability and Independence

  3. Section 5.3Conditional Probability and Independence Learning Objectives After this section, you should be able to… • DEFINE conditional probability • COMPUTE conditional probabilities • DESCRIBE chance behavior with a tree diagram • DEFINE independent events • DETERMINE whether two events are independent • APPLY the general multiplication rule to solve probability questions

  4. Conditional Probability and Independence • What is Conditional Probability? The probability we assign to an event can change if we know that some other event has occurred. This idea is the key to many applications of probability. When we are trying to find the probability that one event will happen under the condition that some other event is already known to have occurred, we are trying to determine a conditional probability. Definition: The probability that one event happens given that another event is already known to have happened is called a conditional probability. Suppose we know that event A has happened. Then the probability that event B happens given that event A has happened is denoted by P(B | A). Read | as “given that” or “under the condition that”

  5. If we know that a person owns a home, what is the probability that the person is a high school graduate? There are a total of 340 people in the sample that own a home. Because there are 221 high school graduates among the 340 home owners, the desired probability is • P(is a high school graduate given owns a home) = 221/340 or 65% Conditional Probability and Independence • Alternate Example: Who Owns a Home • If we know that a person is a high school graduate, what is the probability that the person owns a home? There are a total of 310 people who are high school graduates. Because there are 221 home owners among the 310 high school graduates, the desired probability is • P(owns a home given is a high school graduate) = 221/310 or about 71%

  6. Conditional Probability and Independence • Example: Grade Distributions Consider the two-way table on page 314. Define events E: the grade comes from an EPS course, and L: the grade is lower than a B. Total 6300 1600 2100 Total 3392 2952 3656 10000 Find P(L) Find P(E | L) Find P(L | E) P(L) = 3656 / 10000 = 0.3656 P(E | L) = 800 / 3656 = 0.2188 P(L| E) = 800 / 1600 = 0.5000

  7. Conditional Probability and Independence • Conditional Probability and Independence When knowledge that one event has happened does not change the likelihood that another event will happen, we say the two events are independent. Definition: Two events A and B are independent if the occurrence of one event has no effect on the chance that the other event will happen. In other words, events A and B are independent if P(A | B) = P(A) and P(B | A) = P(B). Example: Are the events “male” and “left-handed” independent? Justify your answer. P(left-handed | male) = 3/23 = 0.13 P(left-handed) = 7/50 = 0.14 These probabilities are not equal, therefore the events “male” and “left-handed” are not independent.

  8. Conditional Probability and Independence • Alternate Example: Allergies • Is there a relationship between gender and having allergies? To find out, we used the random sampler at the United States Census at School website (www.amstat.org/censusatschool) to randomly select 40 US high school students who completed a survey. The two-way table shows the gender of each student and whether the student has allergies. Are the events “female” and “allergies” independent? Justify your answer. P(allergies | female) = 10/23 = 0.435 P(allergies) = 18/40 = 0.45 These probabilities are close but not equal, therefore the events “female” and “allergies” are not independent. Knowing that a student was female slightly lowered the probability that she has allergies.

  9. Conditional Probability and Independence • Tree Diagrams We learned how to describe the sample space S of a chance process in Section 5.2. Another way to model chance behavior that involves a sequence of outcomes is to construct a tree diagram. Consider flipping a coin twice. What is the probability of getting two heads? Sample Space: HH HT TH TT So, P(two heads) = P(HH) = 1/4

  10. Conditional Probability and Independence • Picking Two Sneezers In the previous alternate example, we used a two-way table that classified 40 students according to their gender and whether they had allergies. Here is the table again. Suppose we chose 2 students at random. What is the probability that both students suffer from allergies? To get two students who suffer from allergies, we need to get an allergy sufferer for the first student and an allergy sufferer for the second student. Following along the top branches of the tree, we see that the probability is: P(two allergy sufferers) = P(1st student has allergies and 2nd student has allergies) = P(1st student has allergies) P(2nd student has allergies | 1st student has allergies) = (18/40)(17/39) = 0.196 There is about a 20% chance of selecting two students with allergies.

  11. Conditional Probability and Independence • General Multiplication Rule The idea of multiplying along the branches in a tree diagram leads to a general method for finding the probability P(A ∩B) that two events happen together. General Multiplication Rule The probability that events A and B both occur can be found using the general multiplication rule P(A ∩B) = P(A) • P(B | A) where P(B | A) is the conditional probability that event B occurs given that event A has already occurred.

  12. Conditional Probability and Independence • Example: Teens with Online Profiles The Pew Internet and American Life Project finds that 93% of teenagers (ages 12 to 17) use the Internet, and that 55% of online teens have posted a profile on a social-networking site. What percent of teens are online and have posted a profile? 51.15% of teens are online and have posted a profile.

  13. Conditional Probability and Independence • Example: Playing in the NCAA • About 55% of high school students participate in a school athletic team at some level and about 5% of these athletes go on to play on a college team in the NCAA. What percent of high school students play a sport in high school and go on to play a sport in the NCAA? Almost 3% of high school students will play a sport in high school and in the NCAA.

  14. Example: Who Visits YouTube? See the example on page 320 regarding adult Internet users. What percent of all adult Internet users visit video-sharing sites? P(video yes ∩ 18 to 29) = 0.27 • 0.7 =0.1890 P(video yes ∩ 30 to 49) = 0.45 • 0.51 =0.2295 P(video yes ∩ 50 +) = 0.28 • 0.26 =0.0728 P(video yes) = 0.1890 + 0.2295 + 0.0728 = 0.4913

  15. In a study, 17% of the youth were classified as light media users, 62% were classified as moderate media users and 21% were classified as heavy media users. Of the light users who responded, 74% described their grades as good (A’s and B’s), while only 68% of the moderate users and 52% of the heavy users described their grades as good. According to this study, what percent of young people ages 8-18 described their grades as good? State: What percent of young people get good grades? • Example: Media Usage and Good Grades Do: There are three groups of students who get good grades, those who are light users and get good grades, those who are moderate users and get good grades, and those who are heavy users and get good grades. Because these groups are mutually exclusive, we can add the probabilities of being in one of these three groups. P(good grades) = (0.17)(0.74) + (0.62)(0.68) + (0.21)(0.52) = 0.1258 + 0.4216 + 0.1092 = 0.6566. Plan:If we choose a subject in the study at random, P(light user) = 0.17, P(moderate user) = 0.62, P(heavy user) = 0.21, P(good grades | light user) = 0.74, P(good grades | moderate user) = 0.68, P(good grades | heavy user) = 0.52. We want to find the unconditional probability P(good grades). A tree diagram should help. Conclude:About 66% of the students in the study described their grades as good.

  16. Conditional Probability and Independence • Independence: A Special Multiplication Rule When events A and B are independent, we can simplify the general multiplication rule since P(B| A) = P(B). Definition: Multiplication rule for independent events If A and B are independent events, then the probability that A and B both occur is P(A ∩B) = P(A) • P(B) Example: Following the Space Shuttle Challenger disaster, it was determined that the failure of O-ring joints in the shuttle’s booster rockets was to blame. Under cold conditions, it was estimated that the probability that an individual O-ring joint would function properly was 0.977. Assuming O-ring joints succeed or fail independently, what is the probability all six would function properly? P(joint1 OK and joint 2 OK and joint 3 OK and joint 4 OK and joint 5 OK and joint 6 OK) =P(joint 1 OK) • P(joint 2 OK) • … • P(joint 6 OK) =(0.977)(0.977)(0.977)(0.977)(0.977)(0.977) = 0.87

  17. The First Trimester Screen is a non-invasive test given during the first trimester of pregnancy to determine if there are specific chromosomal abnormalities in the fetus. According to a study published in the New England Journal of Medicine in November 2005 (http://www.americanpregnancy.org/prenataltesting/firstscreen.html), approximately 5% of normal pregnancies will receive a positive result. Among 100 women with normal pregnancies, what is the probability that there will be at least one false positive? State: If 100 women with normal pregnancies are tested with the First Trimester Screen, what is the probability that at least one woman will receive a positive result? Plan: It is reasonable to assume that the test results for different women are independent. To find the probability of at least one false positive, we can use the complement rule and the probability that none of the women will receive a positive test result. P(at least one positive) = 1 – P(no positive results) • Example: First Trimester Screen Do:For women with normal pregnancies, the probability that a single test is not positive is 1 – 0.05 = 0.95. The probability that all 100 women will get negative results is (0.95)(0.95) (0.95) = (0.95)100 = 0.0059. Thus, P(at least one positive) = 1 – 0.0059 = 0.9941. Conclude: There is over a 99% probability that at least one of the 100 women with normal pregnancies will receive a false positive on the First Trimester Screen.

  18. Conditional Probability and Independence • Calculating Conditional Probabilities If we rearrange the terms in the general multiplication rule, we can get a formula for the conditional probability P(B | A). General Multiplication Rule P(A ∩B) = P(A) • P(B | A) P(A ∩B) P(A) P(B | A) Conditional Probability Formula To find the conditional probability P(B | A), use the formula =

  19. Conditional Probability and Independence • Example: Who Reads the Newspaper? In Section 5.2, we noted that residents of a large apartment complex can be classified based on the events A: reads USA Today and B: reads the New York Times. The Venn Diagram below describes the residents. What is the probability that a randomly selected resident who reads USA Today also reads the New York Times? There is a 12.5% chance that a randomly selected resident who reads USA Today also reads the New York Times.

  20. We want to find P(cell phone | landline). Using the conditional probability formula, Conditional Probability and Independence • Example: Who Reads the Newspaper? • In an alternate example in section 5.2, we classified US households according to the types of phones they used. What is the probability that a randomly selected household with a landline also has a cell phone? There is a 77% chance that a landline user also has a cell phone.

  21. Section 5.3Conditional Probability and Independence Summary In this section, we learned that… • If one event has happened, the chance that another event will happen is a conditional probability. P(B|A) represents the probability that event B occurs given that event A has occurred. • Events A and B are independent if the chance that event B occurs is not affected by whether event A occurs. If two events are mutually exclusive (disjoint), they cannot be independent. • When chance behavior involves a sequence of outcomes, a tree diagram can be used to describe the sample space. • The general multiplication rule states that the probability of events A and B occurring together is P(A ∩B)=P(A) • P(B|A) • In the special case of independent events, P(A ∩B)=P(A) • P(B) • The conditional probability formula states P(B|A) = P(A ∩B) / P(A)

  22. Looking Ahead… In the next Chapter… • We’ll learn how to describe chance processes using the concept of a random variable. • We’ll learn about • Discrete and Continuous Random Variables • Transforming and Combining Random Variables • Binomial and Geometric Random Variables

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