Topic:More on Probability

From SharedExperienceProject

Contents 1 Topic Highlights 2 Introduction and Motivation 3 Learning Activities 4 Learning Objectives 5 Topic Notes: Venn Diagrams and the Probability Equations 5.1 What is a Venn Diagram? 5.1.1 Example 1 5.1.2 Example 2 5.1.3 Example 3 5.1.4 Example 4 5.2 How do conditional probabilities fit in? 5.2.1 Example 5 5.3 What about joint probabilities? 5.4 Independence 5.5 Mutual Exclusiveness 5.5.1 Example 6 5.6 The Additive Rule 6 Lecture Notes: Comprehensive Example 6.1 Problem statement 6.1.1 Example 7 6.1.2 Example 8 7 Practice Problems 7.1 Practice Problem 1 7.2 Practice Problem 2

Topic Highlights

(What you will learn)

• Venn diagrams
  
• The relationship between Venn diagrams and contingency tables
• More on basic probability equations
• Using Venn diagrams, contingency tables and the equations to solve probability problems

Introduction and Motivation

(Why learn it)

This topic extends the concepts learned in the topic Introduction to Probability. Same motivation here, just more of the advanced concepts, and not only the case of independent events.

Be sure to understand these too.

Learning Activities

(How the levels of understanding will be gained)

Learning activities for this topic

Type	Name	Direction
Reading	Section 3.3 of Bowerman et al. or: Sections 4.3 and 4.4 of Kvanli et al.	Self-directed
In-class worksheet	Worksheet - More on Probability	Self-directed
In-class discussion	Venn Diagrams and the Probability Equations (below on this page) Comprehensive Example (also below)	Instructor-directed
Practice problems	Practice Problems (below)	Self-directed
Personal activities	Reflection and review of the learning objectives	Self-directed

Learning Objectives

(Levels of understanding to be gained)

Learning objectives for this topic

Level of Understanding	Objective(s)
Very best	Can I solve Practice Problem 2?
Highly satisfactory	Can I solve Practice Problem 1?
Satisfactory	Can I work through Example 7 of the Comprehensive Example on my own? Can I work through Example 8 of the Comprehensive Example on my own? Am I familiar with the equations we used in this topic? Are they on my formula sheet?
Maybe just enough to pass	Can I work through the examples given in the lecture: Example 1, Example 2, Example 3, Example 4, Example 5, and Example 6? Do I understand the Venn diagram concept? Can I draw and interpret one?

Topic Notes: Venn Diagrams and the Probability Equations

These notes are intended to facilitate a discussion introducing the concept of a Venn diagram and to use such diagrams to explain each of the probability equations.

What is a Venn Diagram?

A Venn diagram is a visual way to represent events in a probability situation. Let's have a look at some examples.

Example 1

A manager counts his staff to find a total of 180 people, of which 80 are males.

a) Create the frequency distribution table

b) Draw the Venn diagram

Solution
a) b)

In the above example, be sure you see how the outcomes can only be male or female; the sum of those two areas make up the set of possible outcomes. This is shown explicitly below using shading, just to make sure you've understood this important concept:

From this, it shouldn't be a stretch for you to see that where the complement equation comes from:

1 = P( F ) + P( M )

which is written in the general form as follows:

1 = P( A ) + P( B )

You should also recognize that F and M are complements. The general form of the equation for probabilities of complements is:

Example 2

Remember our marketing manager from last time? She had collected the data below:

Do you recall the contingency table you created for her (just checking):

How would you use the Venn diagram concept to represent the same data for these two classes?

Solution
First, you need to realize that each set of events gets it's own Venn diagram, i.e. one for I and N, and another for U and O: If we put these together (which is normally where we start when studying Venn diagrams), then we get the following:

Notice that we don't usually put any numbers on Venn diagrams - we only use them to represent the classes and the relationships between them. In fact, the usual Venn diagram is even more simple than this one. Because I and N are complements, we don't usually need to label the I. Same goes for U. This gives the standard form of the Venn diagram for two classes of events:

We will see that this conceptual representation can be useful for helping us think about the problems we are posed, and for knowing which equations to use.

Example 3

Can you identify the following areas in the final Venn diagram from Example 2?

a) N and O

b) N or O

Solution

Example 4

What about the following areas?

a) I and O

b) I and U

c) N and U

Solution

How do conditional probabilities fit in?

Now let's use Venn diagrams to understand conditional probabilities and the corresponding equation.

Example 5

Consider the same Venn diagram. How can one use it to think about a conditional probability? For example, let's say we were asked about the probability of respondents not liking the new shoes (N), given that they are older than 18 (O). This is one way to think about it:

From this, it follows that:

P( N | O ) = P( N and O ) / P( O )

The above concept gives rise to the general conditional probability equation given in your textbook:

P( A | B ) = P( A and B ) / P( B )

What about joint probabilities?

The joint probability equation, or multiplicative rule, is hard to show using Venn diagrams. Rather, let's just find it by rearranging the above:

P( A and B ) = P( A | B ) x P( B )

Independence

It is important that you notice that the above equation for joint probability is not the same equation that we saw in the topic Introduction to Probability. The one we studied there was for independent events, whereas the one given above is a general form that always applies.

Two events, A and B are independent if knowing whether A occurs does not change the probability that B occurs.

You should use the form above, unless you are told that the events are independent, or unless you can show that they are using one of the following equations. Tests for independence:

P( A | B ) = P( A )
P( B | A ) = P( B )
P( A and B ) = P( A ) x P( B )

Mutual Exclusiveness

Venn diagrams are also useful for understanding the concept of mutual exclusiveness.

Example 6

Consider the case given below. Notice that it is not possible for an outcome to be both an event of type A and an event of type B. That's because the events are mutually exclusive.

You can test for mutual exclusiveness by checking that:

P( A and B ) = 0

This should make sense to you from the above figure because there is no space where both A and B exist together.

The Additive Rule

The additive rule is a term used to represent the concept of finding the probability that one or another of two events will occur, P( A or B ), but not both.

The Venn diagram in Example 6 is useful to illustrate the additive rule. For the case of mutually exclusive events, you should be able to see that:

P( A or B ) = P( A ) + P ( B )

In other words, the chance of event A or event B is the sum of each chance.

The general case is only slightly more complex. Consider the following Venn diagram where the events are not mutually exclusive:

Here, we want the same area, A + B, but we need to subtract the small shaded area, which will have been counted twice if we don't.

From our earlier discussions, you should recognize this area to be: A and B. So, the general additive rule is:

P( A or B ) = P( A ) + P ( B ) - P ( A and B )

Lecture Notes: Comprehensive Example

The following comprehensive example is intended to facilitate an introduction to problems using the main approaches we've seen so far: Venn diagrams, probability equations and contingency tables.

Problem statement

Imagine that you run the sports centre at Mount Royal College. You carry out a survey of several classes at the college to find out that 20% of the students work out before classes, 50% work out after classes, and 10% (the hard core students) work out both before and after class.

Example 7

a) For the situation provided, define the two relevant classes and give them one-letter labels

Solution to a)
B = students who work out before classes A = students who work out after classes

b) What is the probability corresponding to each class?

Solution to b)
P(B) = 0.20, P(A) = 0.50 We also know enough to write that P(B and A) = 0.10

c) Draw a suitable Venn diagram

Solution to c)

d) What is the probability corresponding to the students who work out before or after class?

Solution to d)
You should recognize this as an "or" problem requiring use of the additive rule: P( B or A ) = P( B ) + P ( A ) - P( B and A ) So, P( B or A ) = 0.20 + 0.50 - 0.10 = 0.60 You should also recognize that this corresponds to the area shaded below:

e) If a student works out before class, what is the probability that he or she will work out after class?

Solution to e)
You should recognize this is a conditional probability question; it doesn't use the word "given," but it was a given that the student works out before class The conditional probability equation applies: P( A | B ) = P( A and B ) / P( B ) = P( B and A ) / P( B ) = 0.10 / 0.20 = 0.50 In other words, if a student works out before class, there is a 50% chance that they will work out after class too

f) If nobody is working out during classes, what is the probability that a student does not work out at all?

Solution to f)
Since we know the probability of students working out (from d) above), this is just the application of the complement equation: P( not working out ) = 1 - P ( working out ) = 1 - P( B or A ) = 1 - 0.60 = 0.40 You should recognize that this corresponds to the complement of the area shown in d):

Example 8

Now let's tackle the same challenge a different way.

a) Draw a contingency diagram for the situation described above, and place the data you have

Setting up the table
First, you need to realize that the table can be set up using class A and its complement, and B and its complement:

Table with given data
Now place the data in the table To do this, you need to arbitrarily assume some size of the sample. Let's use a nice round number like 100:

b) fill in the remaining values, if you haven't already

Complete contingency table

c) Find the answers to questions d), e) and f) of Example 7:

What is the probability corresponding to the students who work out before or after class?
You can find this as follows: P ( B or A ) = 10/100 + 10/100 + 40/100 = 60/100 = 0.60 Or: P( B or A ) = 50/100 + 20/100 - 10/100 = 60/100 = 0.60, which is the same answer as in Example 7 d)

If a student works out before class, what is the probability that he or she will work out after class?
This is a standard conditional probability, just like we studied in the topic Introduction to Probability: P( A | B ) = 10/20 = 0.50, which is the same answer as in Example 7 e)

What is the probability that a student does not work out at all?
This is just: = 40/100 = 0.40, which is the same answer as in Example 7 d)

Practice Problems

As promised last time, this set of problems is more advanced. Try not to look at the solutions until you've worked the problem through.

Practice Problem 1

You are the data analyst for an environmental lobby group and you are concerned about how the upcoming election will impact the chances of a new environmental law going through. Your team has collected the following data from members of the three main political parties about their support for the law:

Find these values:

a) The probability of being for the law and a member of the Conservatives

b) The marginal probability of being against the law

c) The marginal probability of being in the Liberal party

d) The likelihood of a member being against the law given that he or she is an NDP

e) The likelihood of any member not being undecided

f) The probability of being for the law and an NDP

g) P( Liberal ) x P( against | NDP )

Solution

First you need to fill in the rest of the table using the rules about summing in rows and columns: Then you can answer the questions applying what we know about contingency tables: a) P( F and C ) = 82/500 = 0.16 b) P( A ) = 222/500 = 0.44 c) P( L ) = 150/500 = 0.30 d) P( A | N ) = 67/150 = 0.45 e) P( any not U ) = (214 + 222) / 500 = 0.87 f) P( F and N ) = 62/500 = 0.12 g) P( Liberal ) x P( against | NDP ) = 150/500 x 67/150 = 0.13

Practice Problem 2

A recruitment manager surveys 50 potential students of the Bissett School of Business. They are classified on the basis of age and their interest in three of the programs (Entrepreneurship, Supply Chain and Accounting). Of the respondents: 28 were under 20 years old, 11 were between 20 and 25 years old, 20 were interested in Entrepreneurship, 20 were interested in Accounting, 15 of those interested in Entrepreneurship were under 20 years old, 9 of those interested in Supply Chain were under 20, 1 of those interested in Supply Chain is between 20 and 25 years old, and 6 of those interested in Accounting are between 20 and 25 years old.

Part 1: What is the probability of an incoming student being:

a) under 20 years old?

b) interested in Entrepreneurship?

c) interested in Accounting and over 25?

d) over 25 years old, given that he or she is interested in Supply Chain?

e) interested in Supply Chain or under 20?

Part 2: Are being interested in Supply Chain and being over 25 independent events?

HINT: Don't get scared by the written text. Just write down the relevent classes and everything you know. Then go from there.

Solution

Part 1: It is very very difficult to answer this question if you don't realize that a contingency table can be built. Here it is, including everything we've been given: Then you have to fill in the rest of the table knowing the rule about summing the rows and columns: Now it's just a case of applying what we know about contingency tables: a) P( <20 ) = 28/50 = 0.56 b) P( E ) = 20/50 = 0.40 c) P( A and >25) = 10/50 = 0.20 d) P( >25 | S ) = 0.00/10.00 = 0.00 e) P( S or <20 ) = 10/50 + 28/50 - 9/50 = 0.20 + 0.56 - 0.18 = 0.58 Part 2: You can check this using any one of the three equations: This must be true in general: P( A and B ) = P( A ) x P( B ) So, this must be true for our case: P( S and >25 ) = P( S ) x P( >25 ), which we can write as follows using values from the table: 0 = 10/50 x 11/50 Since the left side does not equal the right side, the events are not independent Note: As described in the textbook, if one of the three conditions of independence is found to be true, then they are all true. If one is found to be false, then they are all false.