Topic:Binomial Random Variables

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Contents 1 Topic Highlights 2 Introduction and Motivation 3 Learning Activities 4 Learning Objectives 5 Lecture Notes: Probability Distributions of Binomial Variables 5.1 So, what's the deal? 5.2 So, how does it look? 5.3 So how do I know I'm dealing with the binomial case? 5.3.1 Example 1 5.3.2 Example 2 5.3.3 Example 3 5.4 Computing the probabilities in the table 5.4.1 Example 4 5.4.2 Example 5 5.5 Tables of binomial probabilities? Thank goodness! 5.5.1 Example 6 5.5.2 Example 7 5.5.3 Example 8 6 Lecture Notes: What Else Do I Need to Know? 6.1 Mean of a Binomial Random Variable 6.2 Variance of a Binomial Random Variable 6.2.1 Example 9 7 Practice Problems 7.1 Practice Problem 1 7.2 Practice Problem 2 (ES) 7.3 Practice Problem 3 (ES) 7.4 Other Practice Problems 8 Footnote

Topic Highlights

(What you will learn)

• What a binomial random variable is, and how to recognize one
  
• How a probability distribution table is created for a binomial random variable (although you won't need to do this)
  
• How to look up and use a probability distribution table for a binomial random variable
  
• How to solve probability problems involving binomial random variables
  

Introduction and Motivation

(Why learn it)

In the earlier topic on Discrete Random Variables and Probability Distributions, we learned the basics of what a random variable is and how to create a probability distribution table.

There are a few types of discrete random variables that come up quite often in real life, and the goal of this topic is to look at one in detail: the binomial distribution.

The binomial distribution is especially useful for experiments of a probabilistic nature, in which there are two possible outcomes.

Learning Activities

(How the levels of understanding will be gained)

Learning activities for this topic

Type	Name	Direction
Reading	Section 4.3 of Bowerman et al. or: Section 5.4 of Kvanli et al.	Self-directed
In-class worksheet	Worksheet - Binomial Random Variables	Self-directed
In-class discussion	Probability Distributions of Binomial Variables (below in this page) What Else Do I Need to Know? (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	Can I solve Practice Problem 3?
Highly satisfactory	Can I solve Practice Problem 2? Can I solve Example 4 and Example 6? Can I follow the logic and examples under the heading So, how does it look?
Satisfactory	Can I identify when I am dealing with a binomial random variable? (See the section entitled So how do I know I'm dealing with the binomial case?) Can I solve Example 5, Example 7, Example 8 and Example 9? Do I know how and when to use the equations discussed in this topic? Do I know how and when to use the Binomial Probability tables?
Maybe just enough to pass	Can I solve Example 1, Example 2 and Example 3? Do I understand the concept of a binomial random variable?

Lecture Notes: Probability Distributions of Binomial Variables

These notes are intended to facilitate a discussion on the topics covered in Section 5.4 of Kvanli et al.

So, what's the deal?

The bottom line is this:

1. There are many kinds of random variables out there
2. If you happen to be talking about a binomial random variable (which does happen quite often), then:

• The probability distribution is computed using a very specific formula (see below)
• The formula can be avoided by using the appropriate table in the back of your book (also see below)

3. Everything else is the same, including the general approach to solving problems

So basically, you just need to know the formula (or where to find the values in the table so you can avoid having to compute the formula).

So, how does it look?

When we first looked at probability distribution tables (in the topic Discrete Random Variables and Probability Distributions), we saw the following general form of the probability distribution table:

where X is a discrete random variable (to be defined next), x₁ to x_# make up the set of possible values X can take on, and P(x) is the probability of value x occurring. The total number of possible outcomes is #^[1].

Now, all we're saying is that the probabilities in the right column are given by the following equation:

P(x) = nCx • px • (1-p)n-x

where n is the number of trials in your experiment, and p is the probability of success for each trial (more on these later).

This means the table looks like as follows for the case for an experiment with three trials, i.e. when n=3:

which looks like the following if you compute the values corresponding to the equations on the right (assuming the probability of each outcome, p = 0.50):

For completeness, here are the steps for computing the equation for x=2 for the above case where p = 0.50 and n = 3:

P(x) = _nC_x • p^x • (1-p)^n-x
P(2) = ₃C₂ • (0.5)² • (1-0.5)^3-2
P(2) = 3 • 0.25 • 0.5

P(2) = 0.375

which matches the third value in the table above.

Another way to think of a probability distribution with random variables then, is just as a table of outcomes and the probability of each, where that probability of each is given by the above equation.

So how do I know I'm dealing with the binomial case?

Basically, you know you're dealing with a binomial random variable when the following are true:

1. Your experiment consists of n trials

2. Each trial has 2 mutually exclusive outcomes (usually represented as success and failure)

3. Each trial is independent

4. The probability of the outcome success, usually denoted p, is the same for every trial

5. A random variable can be selected such that: X = number of successes

Let's look at these in more detail with some examples.

Example 1

You flip a coin three times.

a) Check if each of the above conditions passes

b) Can a binomial random variable be used?

Solution

a) Each of the three tosses represents a trial, so n = 3 There are two possible outcomes of each trial (heads and tails) and they are mutually exclusive Each toss of the coin is independent from the one before it The probability of getting either outcome is 0.500, or 50.0% The random variable can be represented as: X = number of heads b) This is sufficient to conclude that it can be represented by a binomial random variable

Example 2

A past survey indicates that 30% of Calgary's population reads the Sun every day. You select 22 people at random from Calgary's population and ask whether they read the Sun.

a) Check if each of the above conditions passes

b) Can a binomial random variable be used?

Solution

a) Asking each person represents a trial, so n = 22 There are two possible outcomes of each trial ("I read it", and "I don't read it") and they are mutually exclusive You selected them randomly, so their responses are independent (NOTE: an example where the responses would not be independent is if you asked each member of the same family - their responses could depend on whether the household subscribes to the Sun, for example) We are told that the probability of a response being "I read it" is 0.300 The random variable can be represented as: X = number of people who read the Sun b) Yes, again

Example 3

Your community association estimates that 40% of its members support a new initiative. You poll 205 members at random to ask whether they will support the initiative.

a) Check if each of the above conditions passes

b) Can a binomial random variable be used?

Solution
a) Each poll taken represents a trial, so n = 205 There are two possible outcomes of each trial ("I do" and "I don't") and these are mutually exclusive Again, you selected them randomly, so they are independent We are told that the probability of a response being "I do" is 0.40 The random variable can be represented as: X = number of people who support b) Yes, again.

Computing the probabilities in the table

We looked at an example earlier of how to compute the values in the probability distribution table by using the equation:

P(x) = _nC_x • p^x • (1-p)^n-x

Example 4

Try your hand at it again for the case where you toss a coin three times. What is the probability of getting three heads?

Solution
First, recognize that n = 3, x = 3 Then you need to know that the probability of getting a head on any toss is p = 0.50 Then you can compute P(3) it as follows: P(x) = nCx • px • (1-p)n-x P(3) = 3C3 • (0.5)3 • (1-0.5)3-3 P(3) = 3 • 0.5 • 0.5 • 0.5 • 1 P(3) = 0.125

Example 5

You might recognize the 3 coin example from our earlier topic on random variables and probability distributions.

a) Take a minute to observe the differences between what you did there and what you did here in Example 4.

b) Why would we want to use the results of the binomial equation anyway? Doesn't it just make things harder?

Solution to a)
There we created a table of outcomes and a table of probabilities in which we calculated the probabilities as the ratio of the frequencies and the total number of values - pretty straightforward and pretty quick Here we have to use the equation for each probability - admittedly a longer process

Solution to b)
We want to use the binomial equation because the probability of each outcome is not always 0.50 for real-world examples - remember that the equation depends on the value of p So, we do need the values given by the equation Luckily, other people have computed those values and they are available in tables - let's take a look

Tables of binomial probabilities? Thank goodness!

As discussed above, there are standard tables from which you can pull the binomial probabilities - without having to compute them!

I hope you see that life just got a whole lot easier. In fact, most problems we see on binomial probability will give you the basic data (n and p) and require you to look up the probabilities in the published tables.

You can find binomial tables in Appendix A, Table A.1 of Kvanli et al. Dig them out now and have a look.

Example 6

Can you use the table find the probabilities corresponding to tossing 3 coins?

Solution
Here: n = 3 again p = 0.50 again Using the table in the textbook, you should be able to find the following From which you should recognize that the probabilities you seek are the ones with the box around them above

Example 7

Now imagine that you have a trick coin, weighted such that it landed on heads 70% of the time.

Can you use the table to find the probabilities now?

Solution
Here: n = 3 again but now, p = 0.70 Using the table in the textbook, you should be able to find the following : From which you should recognize that the probabilities you seek are the ones with the box around them above

Example 8

a) What is the probability of rolling three heads in a row with the weighted coin?

b) How does that compare to the probability of doing the same for an unweighted coin?

Solution
a) Looking at the table for x=3 and P=0.70, P(3) for the weighted coin = 0.343 or 34.3% b) Looking at the table for x=3 and P=0.50, P(3) for the unweighted coin = 0.125 or 12.5% Pretty cool, eh?

Lecture Notes: What Else Do I Need to Know?

There are two more equations you need to be familiar with, and there is no shortcut for these.

Mean of a Binomial Random Variable

The mean of a binomial random variable is given by the following equation:

mean = n • p

where n is the number of trials in your experiment, and p is the probability of a success for each trial.

This is useful for questions related to the expected value.

Variance of a Binomial Random Variable

The other equation is for the variance of a binomial variable:

variance = n • p • (1-p)

where n is the number of trials in your experiment, and p is the probability of a success for each trial.

Example 9

a) What is the expected value for the number of heads obtained by throwing a (normal unweighted) coin three times?

b) The variance?

Solution
n = 3 p = 0.50 a) So, the expected value = mean = 3 x 0.50 = 1.50 b) And the variance = 3 x 0.5 x 0.5 = 0.75 (Notice that you don't need to look up any values in the table - you only need n and p to answer these questions for binomial random variables)

That's pretty much everything you need to know about binomial random variables and how to make (or lookup!) a corresponding probability table. All that's left is to try your hand at some of the types of problems you will see.

Practice Problems

Practice Problem 1

Find the probabilities of the following events for a binomial random variable with n=10 and p=0.5. Use your tables for this.

a) exactly 2 successes

b) more than 2 successes

c) no more than 2 successes

d) less than 2 successes

e) at least 2 successes

Solution
a) P(x=2) = 0.044 b) P(x>2) = 1.0 - P(x<=2) = 1.0 - (0.001 + 0.010 + 0.044) = 0.945 c) P(x<=2) = P(0) + P(1) + P(2) =  0.001 + 0.010 + 0.044 = 0.055 d) P(x<2) = P(0) + P(1) = 0.001 + 0.010 = 0.011 e) P(x>=2) = 1.0 - P(x<2) = 0.989

Practice Problem 2 (ES)

Take the case of the trick coin, weighted such that it landed on heads 70% of the time (recall Example 7).

If you flip it five times:

a) What is the probability of at least 3 heads showing up?

b) What is the probability of getting only one head?

c) What is the probability of getting five tails?

Solution
First, you should recognize that n=5 and that for X=number of heads, p=0.70 Then you should pull the following from the table: Then answer the questions: a) This is the same as asking for the probability of 3 or more = P(3) + P(4) + P(5) = 0.837 b) P(1) = 0.028 c) This is the same as asking for the probability of getting no heads = P(0) = 0.002

Practice Problem 3 (ES)

Richard Branson is a very busy corporate mogul. One of his latest ventures is called Virgin Galactic, which claims to be the world's first spaceline - it will offer normal everyday people the chance to buy a flight into space. Can you imagine?!

Now imagine you have been hired to help him plan his booking system. When finished, each seating class on his spaceships will hold 18 people. Richard's experience in the normal airline business says that 20% of the people who book a flight do not show up for it. He figures that this number will only be 10% for space flights, because they are so expensive. He is thinking of accepting reservations for 20 people for each class.

a) What is the expected number of people who will show up for each class?

b) What is the probability that at least one passenger holding a reservation will not have a seat?

c) What is the probability that it will be exactly two thirds full?

d) What is the probability that there will be exactly 2 empty seats?

e) What is the probability that there will be at least 2 empty seats?

Solution

a) The probability of not showing up is p = 0.10, so the probability of showing up is (1-p) = 0.90 Since n=20, we can compute mean = np = 20 x 0.90 = 18 This might be why he was thinking of accepting 20 people - he only expects 18 to show b) It is easier to solve this problem if you choose X = number who show up In that case, you need to look up the probabilities in the table for 0.90: At least one passenger not having a seat is the same as one passenger or more not having a seat, which would require 19 or 20 people to show up: probability = P(19) + P(20) = 0.270 + 0.122 = 0.392 so, there is a 39.2% chance of this happening - bad for business! c) Two thirds full is 12 people The table says the probability of this is 0.000 or a 0.00% chance, to the level of precision we have been given d) This means that 16 people will show up: probability = P(16) = 0.090, or 9.0% chance e) This means that there are two or more empty seats, so 16 or fewer people will show up: probability = P(16) + P(15) + ... P(1) + P(0) = 0.090 + 0.032 + 0.009 + 0.002 + 0.000 ... = 0.133

Other Practice Problems

See also: Additional problems - Counting and Discrete Random Variables

Footnote

1. ↑ Note that Kvanli et al. is not consistent between Section 5.1 and 5.4. In Section 5.1 it uses n for the number of possible values of X, but in Section 5.4 it uses n for the number of trials. To get around this here, we will represent the number of values of X as # and the number of trials as n.