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