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

## Presentation on theme: "Chapters 14, 15 (part 2) Probability Trees, Odds i)Probability Trees: A Graphical Method for Complicated Probability Problems. ii)Odds and Probabilities."— Presentation transcript:

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

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?

Question 1: What is the probability that a high school baseball player ultimately plays professional baseball P(hs bb player plays professionally) =.061*.094 +.939*.002 =.005734 +.001878 =.007612.061*.094=.005734.939*.002=.001878

Question 2: Given that a high school baseball player played professionally, what is the probability he played in college?.061*.094=.005734.939*.002=.001878 P(hs bb player plays professionally) =.005734 +.001878 =.007612.061*.094=.005734

Example: AIDS Testing zV={person has HIV}; CDC: Pr(V)=.006 zP : test outcome is positive (test indicates HIV present) zN : test outcome is negative zclinical reliabilities for a new HIV test: 1.If a person has the virus, the test result will be positive with probability.999 2.If a person does not have the virus, the test result will be negative with probability.990

Question 1 zWhat is the probability that a randomly selected person will test positive?

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

Probability Tree Multiply branch probs clinical reliability

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

Question 2 zIf 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). zLooks very reliable

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

Summary zQuestion 1: zPr(P ) =.00599 +.00994 =.01593 zQuestion 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

Recap zWe have a test with very high clinical reliabilities: 1.If a person has the virus, the test result will be positive with probability.999 2.If a person does not have the virus, the test result will be negative with probability.990 zBut we have extremely poor performance when the test is positive: Pr(person has HIV given that test is positive) =.376 zIn other words, 62.4% of the positives are false positives! Why? zWhen the characteristic the test is looking for is rare, most positives will be false.

ODDS AND PROBABILITIES zWorld Series OddsWorld Series Odds zFrom probability to odds zFrom odds to probability

From Probability to Odds zIf 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) z 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 Odds to Probability zIf the odds in favor of an event E are a to b, then P(E)=a/(a+b) zIf the odds against an event E are c to d, then P(E’)=c/(c+d) (E’ denotes the complement of E) Team Odds against winning P(E’)=Prob of not winning RED SOX4/14/5=.80 DODGERS5/15/6=.833 TIGERS5/15/6=.833 CARDINALS11/211/13=.846 BRAVES7/17/8=.875 A’s15/215/17=.882 TB RAYS14/114/15=.933 INDIANS14/114/15=.933 REDS16/116/17=.941 PIRATES16/116/17=.941 E = win World Series

Download ppt "Chapters 14, 15 (part 2) Probability Trees, Odds i)Probability Trees: A Graphical Method for Complicated Probability Problems. ii)Odds and Probabilities."

Similar presentations