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# week6 - Lecture 14 Nancy Pfenning Stats 1000 Chapter 7...

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Lecture 14 Nancy Pfenning Stats 1000 Chapter 7: Probability Last time we established some basic definitions and rules of probability: Rule 1: P ( A C ) = 1 - P ( A ). Rule 2: In general, the probability of one event or another occurring is P ( A or B ) = P ( A ) + P ( B ) - P ( A and B ) If the events are mutually exclusive, then P ( A and B ) = 0 and so P ( A or B ) = P ( A ) + P ( B ). Rule 3: In general, the probability of one event and another is P ( A and B ) = P ( A ) P ( B | A ) which we can re-express as P ( B | A ) = P ( A and B ) P ( A ) If the events are independent, P ( B | A ) = P ( B ) and so P ( A and B ) = P ( A ) P ( B ). This time we utilize these rules to solve some more complicated problems, and discuss why at times probabilities can be counter-intuitive. Tree diagrams are often helpful in understanding conditional probability problems. Example (Game Show) Suppose a prize is hidden behind one of three doors A, B, or C. after the contestant picks one of the three, the host reveals one of the remaining two doors, showing no prize. He gives the guest a chance to switch to the door he or she did not select originally. Is the probability of winning the prize higher with the “keep” or “switch” strategy? Use a probability tree to calculate the probability of winning using each strategy. When events occur in stages, we normally are interested in the probability of the second, given that the first has occurred. At times, however, we may want to know the probability of an earlier event having occurred, given that a later event ultimately occurred. Example Suppose that the proportion of people infected with AIDS in a large population is .01. If AIDS is present, a certain medical test is positive with probability .997 (called the sensitivity of the test), and negative with probability .003. If AIDS is not present, the test is positive with probability .015, negative with probability .985 (called the specificity fo the test). Does a positive test mean a person almost certainly has the disease? Let’s figure out the following: if a person tests positive, what is the probability of having AIDS? 58

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A tree diagram is very helpful for this type of problem. We’ll let A and not A denote the events of having AIDS or not, T and not T denote the events of testing positive or not. First we’ll find the overall probability of testing positive. Either a person has AIDS and tests positive or he/she does not have AIDS and tests positive: P ( T ) = P ( A and T ) + P (not A and T ) According to the multiplication rule, it follows that P ( T ) = P ( A ) P ( T | A ) + P (not A ) P ( T | not A ) = . 01( . 997) + . 99( . 015) = . 00997 + . 01485 = . 02482 Now, we can apply the definition of conditional probability to find the probability we seek (prob- ability of having AIDS, given that a person tested positive): P ( A | T ) = P ( A and T ) P ( T ) = . 00997 . 02482 = . 40 Thus, even if a person tests positive, he/she is more likely not to have the disease! . 01 . 99 A not A . 997 . 003 . 985 . 015 T not T T not T P ( A and T ) = . 01( . 997) = . 00997 P (not A and T ) = . 99( . 015) = . 01485 Most students—and even most physicians—would have expected the probability to be much higher.
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week6 - Lecture 14 Nancy Pfenning Stats 1000 Chapter 7...

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