M316 Chapter 12 - M316 Chapter 12 Dr Berg General Rules of...

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Unformatted text preview: M316 Chapter 12 Dr. Berg General Rules of Probability Recall the basic rules of probability. Probability Rules 1 The probability P(A) of any event A satisfies 0 P(A) 1. 2 If S is the sample space in a probability model, then P(S) = 1. 3 Two events A and B are disjoint (or mutually exclusive) if they have no outcomes in common. If A and B are disjoint, then P(A or B) = P(A) + P(B) . This is called the addition rule for disjoint events. 4 (a theorem proved using rules 1, 2, and 3) For any event A P(A does not occur) = 1- P(A) . Independence and the Multiplication Rule Rule 3 gives us the probability that at least one of two events occur in the special case where they are disjoint. Now we examine the probability that two events occur together, also under special circumstances. When we flip a coin twice, or roll two dice, we operate under the assumption that the outcome for one of them has no influence on the other outcome. Multiplication Rule for Independent Events Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent, P(A and B) = P(A)P(B) . This rule can be extended to any number of independent events. Later we will give a more formal definition of independence. Example (12.2) Surviving During WWII, the British found that the probability that a bomber is lost through enemy action on a mission over Europe was 0.05. The probability that the bomber returns safely is 0.95. Let's assume that missions are independent and let Ai be the event of surviving the ith mission. The probability of surviving two missions is P(A1 and A2 ) = P(A1 )P(A2 ) = (0.95)(0.95) = 0.9025 . The probability of surviving three missions is P(A1 and A2 and A3 ) = P(A1 )P(A2 )P(A3 ) = (0.95) 3 = 0.8574 . The probability of surviving twenty missions is P(A1 and A2 and ... and A20 ) = (0.95) 20 = 0.3585 . The tour of duty for an airman was 30 missions. 1 M316 Chapter 12 Dr. Berg Example (12.3) Rapid HIV Testing State: Many people who come to clinics to be tested for HIV don't come back to learn the test results. Clinics now use "rapid HIV tests" that give a result while the client waits. In a clinic in Malawi, for example, use of rapid tests increased the percent of clients who learned their test results from 69% to 99.7%. The tradeoff is that the rapid tests are less accurate than slower laboratory tests. Applied to people who have no HIV antibodies, one rapid test has probability about 0.004 of producing a false positive result. If a clinic tests 200 people who are free of HIV antibodies, what is the chance that at least one false positive will occur? Formulate: It is reasonable to assume that the test results are independent. We have 200 independent events. The probability of not having a false positive is 1- 0.004 = 0.996 . What is the probability that at least one of these events occurs? Solve: The probability is calculated as follows. P(at least one positive) = 1- P(no positives) = 1- P(200 negatives) = 1- (0.996) 200 = 1- 0.4486 = 0.5514 . Conclude: The probability is greater than that at least one of the 200 people will test positive for HIV, even though no one has the virus. Exercise What is the probability of getting at least one ace in a roll of 10 fair dice? The General Addition Rule A Venn diagram illustrates how to modify the addition rule for nondisjoint events. We need to take away the double counting. General Addition Rule for Two Events For any two events A and B, P(A or B) = P(A) + P(B) - P(A and B) . The generalization of this rule is somewhat complicated. For three events, some outcomes may be counted three times and others twice. 2 M316 Chapter 12 Dr. Berg Example (12.4) Motor Vehicle Sales Motor vehicles sold in the United States are classified as either cars or light trucks and as either domestic or imported. "Light trucks" includes SUVs and minivans. "Domestic" means made in North America, so that a Toyota made in Canada counts as domestic. In a recent year, 80% of the new vehicles sold to individuals were domestic, 54% were light trucks, and 47% were domestic light trucks. Choose a vehicle sale at random. Then P(domestic or light truck) = P(domestic) + P(light truck) - P(domestic and light truck) = 0.80 + 0.54 - 0.47 = 0.87 . Thus 87% of vehicles sold were either domestic or light trucks. A vehicle is an imported car if it is neither domestic nor a light truck. Hence P(imported car) = 1- 0.87 = 0.13. Exercise (12.5) Tastes in Music A survey of college students finds that 40% like country music, 30% like gospel music, and 10% like both. Make a Venn diagram and use it to answer these questions. a) What percent of college students like country but not gospel? b) What percent like neither country nor gospel? 3 M316 Conditional Probability Chapter 12 Dr. Berg The probability we assign to an event can change if we know that some other event has occurred. Example (12.5) Trucks Among Imported Vehicles From the Venn diagram of the previous example we can make a twoway table with marginal distributions. Domestic Imported Total Light Truck 0.47 0.07 0.54 Car 0.33 0.13 0.46 Total 0.80 0.20 Suppose we are told that a particular vehicle is imported and asked what the probability is that it is a truck. For this we use the conditional distribution for imported vehicles. In particular, the probability that a vehicle is a truck, given the information that it is imported, is the proportion of trucks in the "Imported" column, 0.07 P(truck | imported) = = 0.35 . 0.20 The vertical bar | can be read as "given the information that" of just "given that." Definition When P(A) > 0 , the conditional probability of B given A is P(A and B) P(B | A) = . P(A) Example (12.6) Imports Among Trucks What is the conditional probability that a randomly chosen vehicle is imported, given that it is a truck? We use the conditional distribution for light trucks. P(imported and truck) 0.07 P(imported | truck) = = = 0.13 P(truck) 0.54 Only 13% of trucks sold are imports. Exercise (12.7) Tastes in Music In the context of exercise 12.5, what is the conditional probability that a student likes gospel music, given that he or she likes country music? The General Multiplication Rule We can use conditional probability to calculate the probability that two events occur when they are not independent. 4 M316 Chapter 12 Dr. Berg General Multiplication Rule For Any Two Events Given events A and B, P(A and B) = P(A)P(B | A) . Example (12.7) Instant Messaging The Pew Internet and American Life Project finds that 87% of teenager (ages 12 to 17) are online, and that 75% of online teens use instant messaging (IM). What percent of teens are online and use IM? P(online and use IM) = P(online)P(use IM | online) = (0.87)(0.75) = 0.6525 So about 65% of teens are online and use IM. The multiplication rule extends to the probability that all of several events occur. The key is to condition each event on the occurrence of all of the preceding events. For example, we have for three events A, B, and C that P(A and B and C) = P(A)P(B | A)P(C | A and B) . Example (12.8) Fundraising by Telephone State: A charity raises funds by calling a list of prospective donors to ask for pledges. It is able to talk with 40% of the names on its list. Of those the charity reaches, 30% make a pledge. But only half of those who pledge actually make a contribution. What percent of the donor list contributes? Formulate: Express the information we are given in terms of events and their probabilities. If A={the charity reaches a prospect} then P(A) = 0.4 . If B={the prospect makes a pledge} then P(B | A) = 0.3 . If C={the prospect makes a contribution} then P(C | A and B) = 0.5 . We want to fine P( A and B and C). Solve: By the general multiplication rule P(A and B and C) = P(A)P(B | A)P(C | A and B) = (0.4)(0.3)(0.5) = 0.06 . Conclude: Only 6% of prospective donors make a contribution. Independence The precise definition of independence is expressed in terms of conditional probability. Definition Two events A and B are independent if P(B | A) = P(B) . Exercise (12.12) Find the probability of being dealt a flush in fivecard draw. Begin by calculating the probability that the first card is a spade, and the second card is a spade, etc. The result is the same for hearts, diamonds and clubs. 5 M316 Tree Diagrams Chapter 12 Dr. Berg A tree diagram can be useful in solving a complicated conditional probability problem. Example (12.9) Adults in the Chat Room State: Online chat rooms are dominated by the young. Let's look only at adult Internet users, age 18 and over. The Pew Internet and American Life Project finds that 29% of adult Internet users are age 18 to 29, another 47% are 30 to 49 years old, and the remaining 24% are age 50 and over. Moreover, 47% of the 18 to 29 age group chat, as do 21% of those aged 30 to 49 and just 7% of those 50 and over. What percent of all adult Internet users chat? Formulate: Restate the facts in terms of probabilities. If we choose an online adult at random, then P(age 18 to 29) = 0.29 P(age 30 to 49) = 0.47 P(age 50 or older) = 0.24 P(chat | age 18 to 29) = 0.47 P(chat | age 30 to 49) = 0.21 P(chat | age 50 and older) = 0.07 . We want to find the unconditional probability P(chat). Solve: We use a tree diagram to help organize the information. We multiply the probabilities down each branch ending in chat and add the results. Thus P(chat)=0.1363+0.0987+0.0168=0.2518. Conclude: About 25% of adult Internet users chat. 6 M316 Chapter 12 Dr. Berg Exercise (12.13) Spelling Errors Spelling errors in a text are either "nonword errors" such as typing "teh" instead of "the", or "word errors" such as typing "lose" instead of "loose". Nonword errors make up 25% of all errors. A human proofreader will catch 90% of nonword errors and 70% of word errors. What percent of all errors will the proofreader catch? 7 ...
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This note was uploaded on 09/14/2009 for the course CH 310 N taught by Professor Blocknack during the Fall '08 term at University of Texas.

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