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
Play coll 0.061
Does not play coll 0.939
Play prof. .094
HS BB Player
.906
Play prof. .002
Does not Play prof. .998
.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?
Play coll 0.061
Does not play coll 0.939
Play prof. .094
HS BB Player
.906
Play prof. .002
Does not Play prof. .998
.061*.094=.005734
.939*.002=.001878
P(hs bb player plays professionally) = .005734 + .001878= .007612
(played in college given that played professionally)
.005734= .7533
.007612
P
.061*.094=.005734
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:
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
What is the probability that a randomly selected person will test positive?
Probability Tree Approach
A probability tree is a useful way to visualize this problem and to find the desired probability.
Probability TreeMultiply
branch probsclinical reliability
clinical reliability
Question 1: What is the probability that a randomly selected person will test positive?
Pr( ) .00599 .00994 .01593P
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
Question 2: If your test comes back positive, what is
the probability that you have HIV? Pr( ) .00599 .00994 .01593P
(have HIV given that test is positive)
.00599= .376
.00599 .00994
P
Summary
Question 1:Pr(P ) = .00599 + .00994 = .01593Question 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 We have a test with very high clinical
reliabilities:1. If a person has the virus, the test result will be
positive with probability .9992. 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.
ODDS AND PROBABILITIES
World Series Odds
From probability to oddsFrom odds to probability
From Probability to Odds
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 Odds to Probability
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, thenP(E’)=c/(c+d)
(E’ denotes the complement of E)
TeamOdds against winning
P(E’)=Prob of not winning
RED SOX 4/1 4/5=.80
DODGERS 5/1 5/6=.833
TIGERS 5/1 5/6=.833
CARDINALS 11/2 11/13=.846
BRAVES 7/1 7/8=.875
A’s 15/2 15/17=.882
TB RAYS 14/1 14/15=.933
INDIANS 14/1 14/15=.933
REDS 16/1 16/17=.941
PIRATES 16/1 16/17=.941
E = win World Series
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