May 08, 2015
6.3 Tree Diagrams
May 9, 2015
Tree diagrams can help simplify some complicated probability problems.
here is an example without a tree diagram first...
Example 6.23 pg 372-373
Only 5% of male high school basketball, baseball, and football players to on
to play at the college level. Of these, only 1.7% enter major league
professional sports. About 40% of the athletes who compete in college and
then reach the pros have a career of more than 3 years.
Events:
A = {competes in college}
B = {competes professionally}
C = {pro career longer than 3 years}
What is the probability that a high school athlete competes in college
and then goes on to have a pro career of more than 3 years?
From the problem we know:
P(A) = 0.05
P(B A) = 0.017
P(C A and B) = 0.4
We want to know:
P(A and B and C) = P(A) P(B A) P(C A and B)
= (0.05)(0.017)(0.4)
= 0.00034
so, 3 in 10,000 high school athletes can expect to compete in college and
have a professional career of more than 3 years.
May 08, 2015
Example 6.24 pg 372-373
Only 5% of male high school basketball, baseball, and football players to on to
play at the college level. Of these, only 1.7% enter major league professional
sports. About 40% of the athletes who compete in college and then reach the
pros have a career of more than 3 years. Additionally, suppose the conditional
probability that a high school athlete reaches professional play given he does
not compete in college is 0.0001.
Events:
A = {competes in college}
B = {competes professionally}
C = {pro career longer than 3 years}
From the problem we know:
P(A) = 0.05
P(B A) = 0.017
P(C A and B) = 0.4
P(B Ac) = 0.0001
What is the probability that a male high school athlete will go on to
professional sports?
this is P(B)
use a tree diagram to organize the information and then answer the question
0.017
college
Pro
B
A
0.05
0.983
Male high
school
athlete
Bc
0.0001
B
0.9999
Bc
0.95
Ac
To answer the question: P(B), first consider the two paths to professional play.
These are disjoint because they cannot happen at the same time (i.e.,can't
go to college and not go to college). Because they are disjoint, we'll eventually
add the probabilities of each of the paths to professional play.
Path 1
P(pro and college) = P(college) P(pro college)
= P(A) P(B A)
= (0.05)(0.017)
= 0.00085
Path 2
P(pro and not college) = P(not college) P(pro not college)
= P(Ac) P(B Ac)
= (0.95)(0.0001)
= 0.000095
so.... P(B) = Path 1 + Path 2
= 0.000945
meaning 9 out of 10,000 male high school athletes will play pro
May 08, 2015
Tree diagrams combine the addition and multiplication rules.
The multiplication rule says that the probability of reaching the end
of any complete branch is the product of the probabilities written
on its segments.
In this example to find the probability of an outcome at the end of a
branch (professional play), we ended up adding the probabilities of all
branches that led to that outcome.