QBM117
Business Statistics
Probability
Probability Trees
Objectives
• To review the rules of probability.
• To introduce probability trees as a means of solving
probability problems.
Rules of Probability
• Complement Rule
P( A )  1  P( A)
• Addition Rule
P( A or B)  P( A)  P( B)  P( A and B)
• Multiplication Rule
P( A and B)  P( A).P( B | A)
Example 1 (Exercise 4.49)
A certain city has a milk and newspaper home
delivery service. During the summer months, it is
estimated that 20% of the households order milk and
60% order newspapers. Of those who order milk.
80% also order newspapers. What proportion of
households
a order milk and newspapers?
b order at most one of the two?
c order neither milk nor newspapers?
M: A household orders milk
N: A household orders newspapers
20% of the households order milk
P(M)=0.20
60% of the households order newspapers
P(N)=0.60
Of those who order milk, 80% also order newspapers
P(N|M)=0.80
a What proportion of households order milk and
newspapers?
P(M and N)  P(M).P( N | M )
 0.20  0.80
 0.16
b What proportion of households order at most one of
the two?
The probability that a household orders at most one
of them is the probability that a household does not
order both milk and newspapers, which is
P(A and B)  1  P(A and B)
 1  0.16
 0.84
c What proportion of households order neither milk nor
newspapers?
We want to find P(M and N)
The event (M or N) consists of all households that
order either milk or newspapers or both.
Hence the event (M and N) = (M or N) consists of all
households that order neither milk nor newspapers.
P( M and N )  P( M or N )
 1  P( M or N )
P( M or N )  P( M )  P( N )  P( N and M )
 0.20  0.60  0.12
 0.64
P( M and N )  1  0.64
 0.36
Example 2
In the evaluation of a sales training program, a firm
found that of 50 persons making a bonus last year,
20 had attended a special sales training program.
The firm has 200 salespersons, 80 of which attended
the sales training program.
What is the probability that a salesperson makes a
bonus given that the salesperson attended the sales
training program?
B = event that a salesperson makes a bonus
B = event that a salesperson does not make a bonus
T = event that a salesperson attended the sales
training program
T = event that a salesperson did not attend the sales
training program
50 of the 200 salespersons made a bonus
50
P( B) 
 0.25
200
Of the 50 salespersons making a bonus, 20 had
attended the sales training program
20
P(T | B) 
 0.4
50
80 of the 200 salespersons attended the sales
training program
80
P(T ) 
 0.4
200
We want to find the probability that a salesperson
makes a bonus given that the salesperson attended
the sales training program.
We want P(B|T).
P ( B and T )
P( B | T ) 
P (T )
0.10

0.40
 0.25
If the firm evaluates the training program in terms of
its effect on the probability that a salesperson makes
a bonus, what is your evaluation of the training
program?
P(B|T)=0.25
=
P(B)=0.25
Hence the probability of making a bonus is not
influenced by whether the salesperson has attended
the sales training program.
The sales training program is not needed.
Probability Trees
• A useful method for calculating probabilities is to use
a probability tree.
• A probability tree is useful for analysing probability
problems in which there are several stages to the
problem.
• Each event is represented by a branch of the tree
with the probability of each event written on the
branch.
• At the end of each branch there is a fork where one
or more branches are drawn representing the next
stage.
How to Draw a Probability Tree
• Read the problem to get an understanding of the
events that are involved.
• Decide on each event that occurs and define it by
giving it a label and writing its definition in full.
• Decide which events occur first. Draw a fork to form
the first stage of your tree, labelling each event at the
end of the branch and writing the associated
probability along the branch.
• At the end of each branch draw a fork to form the
second stage of your tree, an so on.
Joint Probabilities
• Any path through the tree corresponds to one
possible simple event.
• The probability of each simple event can be
calculated by multiplying the probabilities along the
branches that form that path.
Important things to note:
• The sum of the probabilities at each fork of the tree
must add to 1.
• The sum of the joint probabilities must add to 1.
Example 3
A manufacturing firm has plants in Sydney and
Melbourne. The Sydney plant produces 40% of the
total output, with a 10% defect rate. The Melbourne
plant has a 20% defect rate. If a single unit is found
to be defective, is it more likely to have come from
Sydney or Melbourne?
Read the problem to get an understanding of the
events that are involved.
A unit is either produced in Sydney or Melbourne.
A unit is either defective or it is not defective.
Define the events.
S = the event that the unit was produced in Sydney
M = the event that the unit was produced in Melbourne
D = the event that the unit is defective
D = the event that the unit is not defective
What probability are we trying to find?
The questions asked “If a single unit is found to be
defective, is it more likely to have come from Sydney
or Melbourne?”.
We want to find and compare:
the probability that the unit was produced in Sydney
given that it was defective, P(S|D)
and the probability that the unit was produced in
Melbourne given that is was defective, P(M|D)
To calculate P(S|D) we use the formula
P( S and D)
P( S | D) 
P( D)
We need P(S and D) and P(D).
We can calculate these using a probability tree.
To construct a probability tree, we need to decide
which events occur first.
A unit has to be produced before it can be
determined to be defective.
So the first stage of the tree determines where the
unit was produced.
Draw a fork to form the first stage of your tree.
Label each event at the end of the branch.
Write the associated probability along each branch.
P( S )  0.4
P( M )  0.6
S
M
At the end of each branch draw a fork to form the
second stage of your tree, and so on.
P( D | S )  0.1
P( S )  0.4
P( M )  0.6
D
S
P( D | S )  0.9
D
P( D | M )  0.2
D
M
P( D | M )  0.8
D
Calculate the joint probabilities.
P( D | S )  0.1
P( S )  0.4
 0.04
S
P( D | S )  0.9 D
P( D | M )  0.2
P( M )  0.6
D
P(S and D)  0.4  0.1
D
P( S and D )  0.4  0.9
 0.36
P(M and D)  0.6  0.2
 0.12
M
P( D | M )  0.8
D
P( M and D )  0.6  0.8
 0.48
We can now calculate P(D).
P( D)  P( S and D)  P( M and D)
 0.04  0.12
 0.16
We can now calculate P(S|D) and P(M|D).
P ( S and D )
P( S | D) 
P( D)
0.04

0.16
 0.25
P( M | D)  1  P( S | D)
 1  0.25
 0.75
P(S|D)=0.25
Given that a unit is defective, there is a 25% chance
that it was produced in Sydney.
P(M|D)=0.75
Given that a unit is defective, there is a 75% chance
that it was produced in Melbourne.
Hence, if a unit is found to be defective, it is more
likely to have been produced in Melbourne.
Example 4
In Perth 60% of the licensed drivers are 30 years of
age or older, and 40% of the drivers are under 30
years of age. Of all drivers 30 years of age or older,
4% will have a traffic violation in a 12 month period.
Of all drivers under 30 years of age, 10% will have a
traffic violation in a 12 month period.
Assume that a driver has just been charged with a
traffic violation; what is the probability that the driver
is under 30 years of age?
Define the events.
A = the driver is 30 years of age or older
A =the driver is under 30 years of age
B = the driver has a traffic violation
B = the driver does not have a traffic violation
What probability are we trying to find?
We want to find the probability that a driver is under
30 years of age, given that they have been charged
with a traffic violation.
We want to find P(A|B).
To calculate this we use
P( A and B)
P( A | B) 
P( B)
First we need to calculate P(A and B) and P(B).
We can calculate these using a probability tree.
To construct a probability tree, we need to decide on
the stages of the tree.
We will look at a whether a person is aged 30 years
and over.
And then we will look at whether the person has a
traffic violation.
Draw a fork to form the first stage of your tree.
Label each event at the end of the branch.
Write the associated probability along each branch.
P( A)  0.6
P( A )  0.4
A
A
At the end of each branch draw a fork to form the
second stage of your tree, and so on.
P( B | A)  0.04
P( A)  0.6
A
P( B | A)  0.96 B
P( B | A )  0.1
P( A )  0.4
B
B
A
P( B | A )  0.9
B
Calculate the joint probabilities.
P( B | A)  0.04
P( A)  0.6
 0.024
A
P( B | A)  0.96 B
P( B | A )  0.1
P( A )  0.4
B
P( A and B)  0.6  0.04
B
P( A and B )  0.6  0.96
 0.576
P( A and B)  0.4  0.1
 0.04
A
P( B | A )  0.9
B
P( A and B )  0.4  0.9
 0.36
We can now calculate P(A and B) and P(B).
From the probability tree P(A and B) = 0.04 and
P( B)  P( A and B)  P( A and B)
 0.024  0.04
 0.064
Hence P ( A | B )  P ( A and B )
P( B)
0.04

0.064
 0.625
P( A | B)  0.625
Hence the probability that the driver is under 30 years
of age, given that the driver has been charged with a
traffic violation, is 0.625.
Example 5 (Exercise 4.66)
The finance director of a hardware wholesaler has
asked the accountant to ring all customers five days
before their account payments are due, as a means
of reducing the number of late payments. However as
a result of time constraints, only 60% of customers
receive such a call from the accountant. 90% of the
customers called pay on time, while only 50% of
those not called pay on time.
The company has just received a payment on time
from a customer. What is the probability that the
accountant called this customer?
(Answer = 0.73)
Example 6
In examining the borrower characteristics versus loan
delinquency, a bank collected the following
information:
15% of borrowers who have been employed at their
present job for less than 3 years are behind in their
repayments.
5% of borrowers who have been employed at their
present job for at least 3 years are behind in their
repayments.
80% of borrowers have been employed in their
present job for less than 3 years.
1. What is the probability that a randomly selected
account will be for a person who has been in their
present job at least 3 years and who is behind in
making payments?
(Answer = 0.04)
2. What is the probability that a randomly selected
account will be for a person who has been in their
present job less than 3 years or who is behind in
making payments?
(Answer = 0.24)
3. If the account is behind, what is the probability that
the loan is for a person who has been in their
present job for less than 3 years?
(Answer = 0.07)
Reading for next lecture
• Chapter 5 sections 5.1 - 5.3
Exercises
• 4.77