Comm 161
Probability

Agenda


Solving probability problems

11/14/2013

2

Solving Problems
In addition to using the formulas described above, some probability questions can be solved using either trees or tables.
Trees may provide a convenient way to decompose probabilities and help you calculate them.
Marginal

Conditional

Joint

P(B|A)

P(A∩B) = P(A) P(B|A)

=1

P(A)

=1

P(A∩¬B) = P(A) P(¬B|A)

P(¬B|A)

=1
P(B|¬A)

P(¬A)

P(¬A∩B) = P(¬A) P(B|¬A)

=1
P(¬B|¬A)

P(¬A∩¬B) = P(¬A) P(¬B|¬A)
3

Approach to Solving Problems
Here is one approach you can take when solving problems in probabilities: 1. Determine if you will use a tree, probability rules, a table or other method to approach the problem.
2. Define the events of interest.
3. Assign the probability of those events you can determine from the question.
4. Use your chosen technique to find the remaining important probabilities. 4

Sample Problem – Part1
A company has submitted bids on 2 separate contracts. The company president believes that there is a 40% probability of winning the 1st contract. If they win the
1st contract, the probability of winning the 2nd is 70%. However, if they lose the
1st contract, then the probability of winning the 2nd decreases to 50%
- Find the probability that they win only one contract
- Find the probability that they win both contracts

Answer in class / WS

5

Sample Tree Problem
1.
2.
3.

The first step is to identify the events: A – Win First; B – Win Second.
Then determine the given probabilities P(A) = 0.4; P(B|A) = 0.7; P(B|¬A)=0.5
Determine what you need from the tree and fill in values to get it. For part 1 we need P(A∩¬B) + P(¬ A∩B); for part 2, P(A∩B).
Marginal

Conditional

Joint

P(B|A) = 0.70

=1

P(A) = 0.4

=1
P(¬A)

P(A∩B) = P(A) P(B|A)

P(¬B|A)

P(A∩¬B) = P(A) P(¬B|A)

=1
P(B|¬A) = 0.50

P(¬A∩B) = P(¬A) P(B|¬A)

=1
P(¬B|¬A)

P(¬A∩¬B) = P(¬A) P(¬B|¬A)

6

Sample Tree Problem
1.
2.
3.

The first step is to identify the events: A – Win First; B – Win Second.
Then determine the given probabilities P(A) = 0.4; P(B|A) = 0.7; P(B|¬A)=0.5
Determine what you need from the tree and fill in values to get it. For part 1 we need P(A∩¬B) + P(¬ A∩B); for part 2, P(A∩B).
Marginal

Conditional

Joint

P(B|A) = 0.70

=1

P(A) = 0.4

=1
P(¬A) = 0.6

P(A∩B) = P(A) P(B|A)
= (0.4)(0.7) = 0.28

P(¬B|A) = 0.30
P(B|¬A) = 0.50

=1
P(¬B|¬A) =0.50

P(A∩¬B) = P(A) P(¬B|A)
= (0.4)(0.3) = 0.12

=1

P(¬A∩B) = P(¬A) P(B|¬A)
= (0.6)(0.5) = 0.30
P(¬A∩¬B) = P(¬A) P(¬B|¬A)
= (0.6)(0.5)= 0.30

7

Sample Tree Problem – Part 2
Suppose you find out that they did get the second contract, what is the probability that they got the first?

8

Sample Tree Problem
In this case, we now have the tree and we are looking for P(A|B). We know that:

P( A | B) =

P( A  B)
P( A  B)
0.28
=
=
P( B)
P( B | A) P( A) + P( B | ¬A) P(¬A) (0.7)(0.4) + (0.5)(0.6)

Marginal

Conditional

Joint

P(B|A) = 0.70

=1

P(A) = 0.4

=1
P(¬A) = 0.6

P(A∩B) = P(A) P(B|A)
= (0.4)(0.7) = 0.28

P(¬B|A) = 0.30
P(B|¬A) = 0.50

=1
P(¬B|¬A) =0.50

P(A∩¬B) = P(A) P(¬B|A)
= (0.4)(0.3) = 0.12

=1

P(¬A∩B) = P(¬A) P(B|¬A)
= (0.6)(0.5) = 0.30
P(¬A∩¬B) = P(¬A) P(¬B|¬A)
= (0.6)(0.5)= 0.30
9

Sample Table Problem
A firm classifies customer accounts in two ways: whether it is an old customer or a new customer and whether the account is overdue or not. Based on the firm’s 80 accounts, a table was created.
Overdue
New
Old

Not Overdue
5
42

9
24

If one account is selected at random:
a. If the selected account is overdue, what is the probability it is new?
b. If the account is new, what is the probability that it is overdue?
c. Is the age of the account related to whether or not it is overdue, if so how?

10

Sample Table Problem

Your first step should be to