Topic:Discrete Random Variables and Probability Distributions

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Contents 1 Topic Highlights 2 Introduction and Motivation 3 Learning Activities 4 Learning Objectives 5 Lecture Notes: Probability Distributions and Random Variables 5.1 What is a probability distribution? 5.2 What is a random variable? 5.2.1 Example 1 5.3 Steps for creating a probability distribution table 5.3.1 Example 2 5.3.2 Example 3 6 Lecture Notes: Solving these Types of Problems 6.1 Example 4 6.2 Other types of problems 6.2.1 Example 5 6.2.2 Example 6 7 Practice Problems 7.1 Practice Problem 1 7.2 Practice Problem 2 7.3 Practice Problem 3 8 Footnote

Topic Highlights

(What you will learn)

• The concepts of random variables and probability distributions
• What is a discrete random variable
• What is a discrete probability distribution
• How to create discrete probability distributions
• How to solve probability distributions based on the above

Introduction and Motivation

(Why learn it)

In Introduction to Probability and More on Probability, we looked at the basic concept of how to assess the chance of an event of interest occurring among all of the possible outcomes. As we got good at this, we used rules such as joint and conditional probability, and we depended on concepts like the contingency table and Venn diagram. Then in Counting Rules for Probabilities, we figured out how to determine the total number of possible outcomes; not always easy for large numbers of outcomes, but fairly straightforward when you know the concepts of filling slots, permutations and combinations.

So, by now you should be pretty good at figuring out the probability that your event of interest will take place.

But as business analysts, we sometimes need to know the probabilities of more than just one possible outcome. We often want to represent the probabilities of all possible outcomes.

That's where discrete probability distributions come in, and that's the subject of this topic.

Learning Activities

(How the levels of understanding will be gained)

Learning activities for this topic

Type	Name	Direction
Reading	Sections 4.1 and 4.2 of Bowerman et al. or: Section 5.1 of Kvanli et al.	Self-directed
In-class worksheet	Worksheet - Discrete Random Variables and Probability Distributions	Self-directed
In-class discussion	Probability Distributions and Random Variables (below in this page) Solving these Types of Problems (also below)	Instructor-directed
Practice problems	Practice Problems (below in this page)	Self-directed
Personal activities	Reflection and review of the learning objectives (below in this page)	Self-directed

Learning Objectives

(Levels of understanding to be gained)

Learning objectives for this topic

Level of Understanding	Objective(s)
Very best	None[1]
Highly satisfactory	Can I solve Practice Problem 2 and Practice Problem 3? Can I solve Example 6?
Satisfactory	Can I solve Practice Problem 1? Can I solve Example 4, and Example 5? Do I know how and when to use the two equations given under Other types of problems? Do I know whether a random variable is numeric or not? Am I familiar with the steps for creating a probability distribution table?
Maybe just enough to pass	Can I solve Example 1, Example 2 and Example 3? Do I understand the difference between a random variable and a probability distribution?

Lecture Notes: Probability Distributions and Random Variables

These notes are intended to facilitate an introductory discussion on the topic of probability distributions and random variables.

What is a probability distribution?

Basically, a probability distribution is just a better (i.e. more complete) way to represent the various outcomes of an experiment and their probabilities.

Generally speaking, we can write a probability distribution as follows:

where X is a discrete random variable (to be defined next), x₁ to x_n make up the set of possible values X can take on, and P(x) is the probability of value x occurring. As usual, the total number of possible outcomes is n.

Another way to think of a probability distribution then, is just as a table of outcomes and the probability of each.

What is a random variable?

A random variable is just a numerical representation of the outcomes of an experiment.

Let's try to understand random variables and probability distributions using an example we've seen many times before.

Example 1

Let's consider the now-familiar experiment of tossing two coins simultaneously.

a) How many possible outcomes are there (use intuition or the slot filling concept we saw in Counting Rules for Probabilities)?

Solution
total number of outcomes = n1 x n2 = 2 x 2 = 4

b) What are the possible outcomes?

Solution
We've seen these before: H, H H, T T, H T, T

c) Can you think of some possible events of interest?

Solution
You might be interested in the following and others: The number of heads in a roll The number of tails in a roll Whether a roll has no tails Whether both coins are tails

d) Now assume that the event of interest is "the number of heads rolled each time". Can you build a table containing with the following column headings: 1) "outcome" 2) "number of heads" rolled per outcome?

Solution

e) Now let's define a random variable X such that it corresponds to the number of heads in a roll and such that the possible values are 2, 1 and 0. Can you build a probability distribution table with the following column headings: 1) "x", and 2) "frequency, m", and 3) "P(x)"?

Solution

There! You just defined a discrete random variable and created your first probability distribution table. Now let's take a look at the steps we followed.

Steps for creating a probability distribution table

The basic steps are as follows:

1. Define the random variable, X = the event of interest
2. Create a table of outcomes with the two headings: "outcomes", and "value of X"
   • List all of the possible outcomes and the values of X that correspond to each
3. Create probability distribution table with the three headings: x, "frequency, m", and P(x)
   • Be sure to only list each x once
   • Put the totals for the frequency and probability columns

Now, let's try some examples.

Example 2

Imagine that you flip three coins simultaneously.

a) Create a table of outcomes for the random variable X=number of heads rolled

b) Build the corresponding probability distribution table

Solution to a): Table of outcomes

Solution to b): Probability distribution table

Example 3

Let's do another. Now imagine that you roll two 4-sided dice simultaneously.

a) Create a table of outcomes for the random variable X="sum of both dice"

b) Build the corresponding probability distribution table

Solution to a): Table of outcomes

Solution to b): Probability distribution table

Lecture Notes: Solving these Types of Problems

Okay, so now you know how to create a probability distribution table. Very important, but not usually the whole deal. Let's take a look at some of the types of problems you will see.

Example 4

Could you answer the following questions from scratch?

a) When rolling two 4-sided dice, what sum of both are you most likely to obtain?

b) What is the probability of obtaining it?

Solution
You should see from Example 3, that: a) The sum with the highest probability of occurring is 5 b) The probability of obtaining it is 0.25

Other types of problems

After building a probability distribution table, we are often faced with having to use it to answer some questions.

For discrete probabilities, there are two types of probability questions you will often see, both of which have to do with the probability of more than one event occurring.

The first comes from the equation for joint independent events:

P(A and B and C) = P(A) • P(B) • P(C)

The second comes from the additive ("or") equation for mutually exclusive events:

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

You can learn more about these in Section 4.6, but most of what you need is shown by the following examples.

Example 5

Given the following probability distribution table where X=sum of both dice, what is the probability of rolling the dice such that the sum is less than 5?

Solution
You might be able to see this intuitively. Otherwise you can use the equation for mutually exclusive events given above: P(x<5) = P(x=2) + P(x=3) + P(x=4) = 1/16 + 2/16 + 3/16 = 6/16 = 0.3750

Example 6

You're in charge of a hamburger stand at the Calgary Stampede. Assume that 5 out of every 100 of your burgers is cooked imperfectly (they come out slightly raw). An inspector from a local meat inspection company, The Cowpokes, is going to inspect three of your burgers and will only certify your stand if all three are perfect. What is the probability of passing the inspection?

Solution
You should use the equation for joint independent events given above: P(perfect) = 1.0 - P(imperfect) = 1.0 - 5/100 = 0.950 P(pass) = P(perfect) x P(perfect) x P(perfect) = 0.950 x 0.950 x 0.950 = 0.857, or 85.7% You probably want to learn how to cook them a little longer

The kind of problem shown in the last example can also be posed using a probability distribution table. Check out the Practice Problems, for example

Practice Problems

Practice Problem 1

In a batch of circuit boards, there are two boards that need to be returned to the factory, three boards that need repair but do not need to be returned, and three boards that are in good working condition.

a) Follow the steps for creating a probability distribution table for this situation. For the random variable, use values of 0, 1 and 2 for the outcomes return, repair and good, respectively.

b) What is the probability that a circuit board selected at random needs to be returned to the factory?

c) What is the probability that a circuit board is in good working order?

Solution
a) b) 0.250 c) 0.375

Practice Problem 2

If an inspector is going to decide whether to certify a shipment from Problem 1 by inspecting three circuit boards, what are the chances he will do so if all three boards need to be in good condition to justify certification?

Solution
This is similar to Example 6 P(certify) = P(good) x P(good) x P(good) = 0.375 x 0.375 x 0.375 = 0.053 In other words, there is only a 5.3% chance of him certifying the shipment

Practice Problem 3

a) Repeat Example 3 where the random variable, X=average of both dice.

b) What is the probability of rolling such that the average is 3.0 or higher?

Solution
a) b) P(x >= 3) = P(x=3) + P(x=3.5) + P(x=4) = 3/16 + 2/16 + 1/16 = 6/16 = 0.375 = 37.5%

Footnote

1. ↑ No really tough problems are given here. They will be given in later sections on probability distributions - we're just learning the basics in this topic.