1. + 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

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

3. + Section 5.3 Conditional Probability and Independence 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 Learning Objectives

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 ofprobability. 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. + Conditional Probability and Independence Alternate Example: Who Owns a Home High School Graduate? YesNoTotal Homeowner221119340 Not a Homeowner8971160 Total310190500 1. If we know that a person owns a home, what is the probability that theperson is a high school graduate? There are a total of 340 people in thesample that own a home. Because there are 221 high school graduatesamong the 340 home owners, the desired probability is P (is a high school graduate given owns a home) = 221/340 or 65% 2. If we know that a person is a high school graduate, what is the probabilitythat the person owns a home? There are a total of 310 people who arehigh school graduates. Because there are 221 home owners among the310 high school graduates, the desired probability is P (owns a home given is a high school graduate) = 221/310 or about 71%

6. + 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. Conditional Probability and Independence Find P(L) Find P(E | L) Find P(L | E) Total 3392 2952 3656 10000 Total 6300 1600 2100 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 When knowledge that one event has happened does not change the likelihood that another event will happen, we say the twoevents are independent. Conditional Probability and Independence 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). 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. Are the events “male” and “left-handed” independent? Justify your answer. Example:

8. + Alternate Example: Allergies Is there a relationship between gender and having allergies?To find out, we used the random sampler at the United StatesCensus at School website ( www.amstat.org/censusatschool ) to randomly select 40 US high school students who completeda survey. The two-way table shows the gender of eachstudent and whether the student has allergies. www.amstat.org/censusatschool Conditional Probability and Independence P(allergies | female) = 10/23 = 0.435P(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. Are the events “female” and “allergies” independent? Justify your answer. FemaleMaleTotal Allergies10818 No Allergies13922 Total231740

9. + Tree Diagrams We learned how to describe the sample space S of a chance process in Section 5.2. Another way to model chancebehavior that involves a sequence of outcomes is to constructa tree diagram. Conditional Probability and Independence 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. + Picking Two SneezersIn the previous alternateexample, we used a two-waytable that classified 40 studentsaccording to their gender andwhether they had allergies.Here is the table again. Conditional Probability and Independence 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(1 st student has allergies and 2 nd student has allergies) = P(1 st student has allergies) P(2 nd student has allergies | 1 st student has allergies) = (18/40)(17/39) = 0.196 There is about a 20% chance of selecting two students with allergies. FemaleMaleTotal Allergies10818 No Allergies13922 Total231740

11. + 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. Conditional Probability and Independence 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. General Multiplication Rule

12. + 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 profileon a social-networking site. What percent of teens are online and have posted a profile? Conditional Probability and Independence 51.15% of teens are online and have posted a profile.

13. + Example: Playing in the NCAA About 55% of high school students participate in a school athletic team atsome level and about 5% of these athletes go on to play on a college team inthe NCAA. What percent of high school students play a sport in high school and go on to play a sport in the NCAA? Conditional Probability and Independence 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 ∩ 18 to 29) = 0.27 0.7 =0.1890 P(video yes ∩ 30 to 49) = 0.45 0.51 =0.2295 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 ∩ 50 +) = 0.28 0.26 =0.0728 P(video yes) = 0.1890 + 0.2295 + 0.0728 = 0.4913

15. + Example: Media Usage and Good Grades State: What percent of young people get good grades? 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 unconditionalprobability P (good grades). A tree diagram should help. 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 areheavy users and get good grades. Because these groups are mutually exclusive, we canadd 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. Conclude: About 66% of the students in the study described their grades as good. In a study, 17% of the youth were classified as light media users, 62% were classified asmoderate media users and 21% were classified as heavy media users. Of the light userswho responded, 74% described their grades as good (A’s and B’s), while only 68% of themoderate users and 52% of the heavy users described their grades as good. According tothis study, what percent of young people ages 8-18 described their grades as good?

16. + 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 ). Conditional Probability and Independence 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) 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 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? Example:

17. + Example: First Trimester Screen 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 positiveresult? 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 usethe complement rule and the probability that none of the women will receive apositive test result. P (at least one positive) = 1 – P (no positive results) 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 negativeresults 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. The First Trimester Screen is a non-invasive test given during the first trimester ofpregnancy 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 normalpregnancies, what is the probability that there will be at least one false positive? http://www.americanpregnancy.org/prenataltesting/firstscreen.html

18. + 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 ). Conditional Probability and Independence P(A ∩ B) = P(A) P(B | A) General Multiplication Rule To find the conditional probability P(B | A), use the formula Conditional Probability Formula P(A ∩ B)P(B | A)P(A)P(A) =

19. + 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 ? Conditional Probability and Independence There is a 12.5% chance that a randomly selected resident who reads USA Today also reads the New York Times.

20. + Example: Who Reads the Newspaper? In an alternate example in section 5.2, we classified US households accordingto the types of phones they used. What is the probability that a randomly selected household with a landline also has a cell phone? Conditional Probability and Independence There is a 77% chance that a landline user also has a cell phone. Cell Phone No Cell Phone Total Landline0.600.180.78 No Landline 0.200.020.22 Total0.800.201.00 We want to find P (cell phone | landline). Using the conditional probabilityformula,

21. + Section 5.3 Conditional Probability and Independence 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) Summary

22. + Looking Ahead… 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 In the next Chapter…