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
Play prof. .094
.061*.094=.005734
Play coll 0.061
.906
HS BB Player
Play prof. .002
Does not play coll
0.939
Does not Play
prof. .998
.939*.002=.001878
P(hs bb player plays professionally)
= .061*.094 + .939*.002
= .005734 + .001878
= .007612
Question 2: Given that a high school baseball player played
professionally, what is the probability he played in college?
Play prof. .094
Play coll 0.061
.906
.061*.094=.005734
P(hs bb player plays professionally)
= .005734 + .001878
= .007612
HS BB Player
Play prof. .002
.939*.002=.001878
Does not play coll
0.939
Does not Play
prof. .998
P(played in college given that played professionally)
.005734
=
 .7533
.007612
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 Tree
clinical
reliability
clinical
reliability
Multiply
branch probs
Question 1: What is the probability that a
randomly selected person will test positive?
Pr( P )  .00599  .00994  .01593
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( P )  .00599  .00994  .01593
P (have HIV given that test is positive)
.00599
=
 .376
.00599  .00994
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
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 .999
2. 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 odds
From 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, then
P(E’)=c/(c+d)
(E’ denotes the
complement of E)
E = win World Series
Team
Odds 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