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Chapters 14, 15 (part 2) Probability Trees, Odds

Chapters 14, 15 (part 2) Probability Trees, Odds. Probability Trees: A Graphical Method for Complicated Probability Problems. Odds and Probabilities. Probability Tree Example: probability of playing professional baseball.

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Chapters 14, 15 (part 2) Probability Trees, Odds

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  1. Chapters 14, 15 (part 2) Probability Trees, Odds Probability Trees: A Graphical Method for Complicated Probability Problems. Odds and Probabilities

  2. Probability Tree Example: probability of playing professional baseball • 6.1% of high school baseball players play college baseball. Of these, 9.4% will play professionally. • Unlike football and basketball, high school players can also go directly to professional baseball without playing in college… • studies have shown that given that a high school player does not compete in college, the probability he plays professionally is .002. Question 1: What is the probability that a high school baseball player ultimately plays professional baseball? Question 2: Given that a high school baseball player played professionally, what is the probability he played in college?

  3. Question 1: What is the probability that a high school baseball player ultimately plays professional baseball .061*.094=.005734 .939*.002=.001878 P(hs bb player plays professionally) = .061*.094 + .939*.002 = .005734 + .001878 = .007612

  4. Question 2: Given that a high school baseball player played professionally, what is the probability he played in college? .061*.094=.005734 .061*.094=.005734 P(hs bb player plays professionally) = .005734 + .001878 = .007612 .939*.002=.001878

  5. Example: AIDS Testing • V={person has HIV}; CDC: Pr(V)=.006 • P : test outcome is positive (test indicates HIV present) • N : test outcome is negative • clinical reliabilities for a new HIV test: • If a person has the virus, the test result will be positive with probability .999 • If a person does not have the virus, the test result will be negative with probability .990

  6. Question 1 • What is the probability that a randomly selected person will test positive?

  7. Probability Tree Approach • A probability tree is a useful way to visualize this problem and to find the desired probability.

  8. Probability Tree Multiply branch probs clinical reliability clinical reliability

  9. Question 1: What is the probability that a randomly selected person will test positive?

  10. Question 2 • If your test comes back positive, what is the probability that you have HIV? (Remember: we know that if a person has the virus, the test result will be positive with probability .999; if a person does not have the virus, the test result will be negative with probability .990). • Looks very reliable

  11. Question 2: If your test comes back positive, what is the probability that you have HIV?

  12. Summary • Question 1: • Pr(P ) = .00599 + .00994 = .01593 • Question 2: two sequences of branches lead to positive test; only 1 sequence represented people who have HIV. Pr(person has HIV given that test is positive) =.00599/(.00599+.00994) = .376

  13. Recap • We have a test with very high clinical reliabilities: • If a person has the virus, the test result will be positive with probability .999 • If a person does not have the virus, the test result will be negative with probability .990 • But we have extremely poor performance when the test is positive: Pr(person has HIV given that test is positive) =.376 • In other words, 62.4% of the positives are false positives! Why? • When the characteristic the test is looking for is rare, most positives will be false.

  14. ODDS AND PROBABILITIES World Series Odds From probability to odds From odds to probability

  15. If event A has probability P(A), then the odds in favor of A are P(A) to 1-P(A). It follows that the odds against A are 1-P(A) to P(A) If the probability the Boston Red Sox win the World Series is .20, then the odds in favor of Boston winning the World Series are .20 to .80 or 1 to 4. The odds against Boston winning are .80 to .20 or 4 to 1 From Probability to Odds

  16. If the odds in favor of an event E are a to b, then P(E)=a/(a+b) If the odds against an event E are c to d, then P(E’)=c/(c+d) (E’ denotes the complement of E) From Odds to Probability E = win World Series

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