Theoretical vs. Empirical Probabilities

MAT 121 Spring 2013 Fisher

Sections Covered: 5.5; 6.1-6.3

In the previous class, we discussed a wide variety of topics regarding probability:

Theoretical vs. Empirical Probabilities

o Theoretical Probability:

o Empirical Probability:

Disjoint/Mutually Exclusive Events (or)

o

Independent Events (and)

o

Important: If the second event is not independent then we must adjust the

probability accordingly (for instance, if there is replacement versus when there is

not replacement)

Conditional Probability

o

One final IMPORTANT piece of the probability puzzle is the ability to count potential outcomes.

Especially when talking about theoretical probabilities, potential outcomes are literally half the

equation.

There are multiple counting techniques and which one we should use depends entirely on several

factors. Some of those factors include whether repetition is allowed or whether order matters. We

will now explore all the counting methods and how to tell when to use them.

The first technique we will discuss is the most basic and fundamental method, it is actually referred to

as the Fundamental Counting Principle. The Fundamental Counting Principle, FCP, states that to find

all possible outcomes you simply multiply the possible choices for each event. The product of this

process represents the total possible outcomes or arrangements.

For example, if a menu gave you a choice of 3 appetizer, 4 entrees, and 5 desserts, then there are:



The text will refer to this as the Multiplication Rule of Counting, stating that if you have p selections

for the first choice, q selections for the second choice, r selections for the third choice, and so on, then

the task of making these selections can be done in different ways.

There are many ways you can be asked to utilize this principle, below are a few samples of how the

multiplication rule of counting can be implemented.

Example 1: Employees at a local business are issued employee ID’s consisting of 2 letters and 3

numbers. The letters may repeat however the numbers may not. How many different employee ID’s

are possible using this system?

Note: In this example, some repetition was allowed (for the letters) but some choices could not be

used more than once (for the numbers). When using the multiplication rule of counting, you must pay

attention to whether repetition is allowed since that will impact how many choices you have for

subsequent selections.

Example 2: A student body council must select a president, vice-president, and secretary for their

group. If there are 50 members of the council, how many different ways could the officers be

selected?

Sometimes when using the multiplication rule of counting you will be asked to use all the choices. For

instance, if I wanted to arrange 6 textbooks on a shelf, how many ways could I position them?

You would have 6 choices for the first position, 5 choices for the second position, 4 choices for the

third, 3 choices for the fourth, 2 choices for the fifth, and lastly only 1 choice for the last position.

Using the rule, we would multiply . If we know we need to use all

the choices, we can save time using a special symbol known as the factorial symbol (n!). Using the

factorial symbol (!), we know to multiply the number n by all values in descending order down to 1.

Using the example from above regarding the 6 textbooks, we really were calculating:



An example that utilizes factorial:

A traveling sales representative must travel to 8 different locations to service existing accounts. How

many different routes are possible?

As discussed above, sometimes repetition is not allowed. There is a special name for the multiplication

rule for counting when it involves making distinct selections, this name is permutations. The book

defines permutations as “an ordered arrangement in which r objects are chosen from n distinct objects

so that and repetition is not allowed.”

The symbol for permutation is nPr.

There are 3 requirements to consider when using the permutation function:

1) Are the objects distinct?

2) Is repetition prohibited?

3) Is order important?

If all 3 answers are yes, then we use the permutation function: nPr =

Before we use it to count possible outcomes, let’s practice using the formula first.

Calculate 5P3

5P3 =

Note: We stop expanding at 2!. Why?

Calculate 6P6

6P6 =

Important!



Now let’s use permutations within the context of a counting outcomes problem.

Example: Are you feeling lucky??? In horse racing, one can bet on which horses will finish in 1st, 2nd,

and 3rd. also known as the trifecta. If there are 10 horses in the race, how many ways can the first 3

positions be filled?

Since order matters in a race, there can be no repeats (a horse can’t get 1st and 2nd place), and there

are 10 different horses racing, this is a perfect time to use a permutation.

There are 10 choices (n) and we want to pick 3 (r), so our permutation will be:

10P3 =

So there are 720 different ways that 10 horses could finish in the first 3 positions. No one said

gambling was easy!

Try this! Going one step further with the previous example, a perfecta is a term used in horse racing

where a gambler attempts to pick who the top 4 finishers will be. How many ways can a 10 horse race

finish for the top 4 horses?

10P4 =

You can start to see why this is so difficult to do, however if you can catch lightning in a bottle you

could become quite wealthy!

DISCLAIMER!!! I DO NOT IN ANY WAY ENCOURAGE STUDENTS TO PARTICIPATE IN GAMBLING LEGAL OR OTHERWISE!!!

Let’s think back to the study body council from above, instead this time let’s choose 5 members to

serve on a fundraising committee. Does order matter anymore? The answer is no. It doesn’t matter

what order I choose the 5 members since there is no inherent difference between committee

members in terms of power or responsibilities. Since order no longer matters, we must find another

method (permutation cannot be used if order does not matter) for determining the total number of

ways a 5 member committee could be chosen. Enter combinations! Combinations are very similar to

permutations (there must be n distinct objects and repetition is not allowed) with one significant

difference, order no longer matters.

The formula for combinations is nCr =

Note: Look how similar the formulas are!



Let’s just practice calculating a combination.

Calculate 6C2

6C2 =

Calculate 8C8

8C8 =

Let’s put combinations into context.

Example A local charity has raised funds to send some local honor society students to Washington, D.C.

for a national honor society convention. There are 20 members in the local chapter and they must

select 5 members to send. How many different group arrangements are possible?

20C5 =

One of the skills we worked on earlier this semester was selecting samples from populations. Using

combinations, you can determine how many samples of size n can be selected from a population of

size N. For instance, how many samples of size n = 3 can be selected from a population of 15

individuals?

15C3 =

Up to this point we have made one common assumption for both permutations and combinations (and

the multiplication rule for counting in general), that all choices involved were distinct or all different.

What happens when we take this assumption away, what if we do have choices that repeat?

Think about this: How many ways could the letters in the word STATISTICS be rearranged?



The problem here, unlike in the previous counting methods, is there are letters that are used more

than once. The letter S appears 3 times, T 3 times, and I twice. So if I randomly choose a T, which T did

I pick? This is what is meant by nondistinct objects.

So if I want to rearrange the letters in the word STATISTICS, we must consider each letters potential

sequence within the entire group of choices.

So there are:

10C3 ways of positioning the S’s

o (Notice we use a combination since we can’t tell the difference between S’s)

7C3 ways of positioning the T’s

o (Notice the number of choices, n, is decreasing since we have less positions left)

4C2 ways of positioning the I’s

2C1 ways of positioning the A’s

1C1 ways of positioning the C’s

So we would “simply” multiply: (10C3) (7C3) (4C2) (2C1) (1C1) =

Easy right??? With a few cancellations the product becomes nicer,

. We will discuss a more

direct approach, but first let’s finish the calculation:

We used combinations above because the choice of the each letter (especially the nondistinct ones)

didn’t matter since we could not tell the difference between each of the S’s, T’s, or I’s. Overall though,

the order of how we arrange the letters DOES matter and therefore this is a permutation. Luckily there

is a formula for permutations with nondistinct objects:

The number of permutations of n (nondistinct) objects =

Try This! How many ways can the letters in the word MISSISSIPPI be rearranged?

There are 11 total objects (n) with: 1 M (n1), 4 I’s (n2), 4 S’s (n3), and 2 P’s (n4).

Total Arrangements =



Let’s put all this counting techniques to use, let’s calculate some probabilities.

Winning the New York State lottery, everyone’s dream and the easy way to the high life. But what are

the chances? You may have been told that you have a better chance of being struck by lightning.

According the National Weather Service, the chances of being struck by lightning in any given year are

. These are pretty long odds and therefore highly unlikely. On a side note, there is

approximately a

chance of being struck by lightning in your lifetime and scarier yet

odds of

knowing someone who will be struck. That means, statistically speaking, if you have 1,000 Facebook

friends that 1 of them will get struck in your lifetime! Yikes!!!

Back to the problem at hand, is winning the NY Lottery less likely than getting struck by lightning during

any given year?

First we must determine how many ways are there of having the 6 numbers on your ticket. There are

59 numbers in play during a lottery drawing and since the order of the numbers doesn’t matter we will

calculate the combination 59C6.

59C6 =

Since you get two plays for $1, the number of ways of achieving success is 2. In other words, you have

two chances per ticket.

This is approximately

. In summary, 1 out of 1 million people get struck by

lightning while 1 out of 25 million wins the lottery. Your odds of getting struck by lightning are WAY

better than winning the lottery. As a matter of fact, you are almost more likely to get struck by

lightning AND know someone else who got struck by lightning than you are to win the lottery. But

someone has to win, $1 and a dream! Again, see disclaimer above.



We now move into chapter 6. We will review some terminology from earlier this semester.

Discrete

Continuous

Random

Using these ideas, we will now expand on them.

A random variable is a numerical measure of the outcome of a probability experiment, so its value is

determined by chance. A random variable can be either discrete or continuous.

A discrete random variable is a random variable that has either a finite or countable number of values.

A continuous random variable is a random variable that has infinitely many values.

A discrete random variable can be plotted on a number line with spaces between each point whereas a

continuous random variable can be plotted on a number line in an uninterrupted way.

If I rolled a standard number cube 5 times and the random variable X represents the number of 3’s that

resulted from the rolls, then the possible values of X are x = 0, 1, 2, 3, 4, or 5. Therefore X in this case

would be a discrete random variable.

If I was measuring the wait time of the next car to pull through the drive-thru window at a fast-food

restaurant and the random variable X represents the wait time, then the possible values of X are

infinite (x > 0 minutes). Therefore X in this case would be a continuous random variable.

We will spend the majority of our remaining time focusing on the discrete random variable.

A probability distribution is similar to a frequency distribution. A probability distribution of a discrete

random variable lists all possible values of the random variable along with their corresponding

probabilities. Like frequency distribution, probability distributions can be modeled in the form of a

table, graph, or formula.



The data below represents the number of free throws made (out of 2 shots). Use the results to create

a probability distribution for the discrete random variable X, the number of free throws made.

x P(x)

0

1

2

As you are analyzing and/or creating probability distributions, keep 2 things in mind:

1) All probabilities must be between 0 and 1

2) All probabilities must add up to 1.

Hopefully you are realizing how similar this process is to what we did earlier this semester with

frequency distributions. Just like before, we can also display these distributions in graph form,

especially histograms. Probability histograms are excellent visual aids for helping convey the

likelihood of particular (in this case, discrete) events. Below I have included a probability histogram

for the discrete random variable X, free throws made.

1 1 0 1 1

1 0 1 2 1

2 2 1 1 1

1 2 1 0 1

2 2 0 1 1



Keeping with the theme of similarity, we will now discuss the how to find the center of the discrete

random variable. Recall when we found mean of data, we used the observed value to calculate the

statistics/parameters. We will do something similar here, however rather than using observed

outcomes we will use each random variable and its corresponding probability to determine the mean.

In order to find the mean of a discrete random variable we use the formula is . In

words, we simply multiply each possible value of the random variable by its probability and then add

the products together. Let’s practice using this formula using the example above.

x P(x) x * P(x)

0

0 * 0.16 = 0

1

1 * 0.6 = 0.6

2

2 * 0.24 = 0.48

So . What does this mean? We would expect the player to make 1.08 out

of 2 shots from the free throw line.

Since the mean of the discrete random variable is what we expect to happen given a set of

circumstances, we also refer to the mean as the expected value.

Practice with Expected Value

Shawn and Maddie purchase a foreclosed property for $50,000 and spend an additional $27,000 fixing

up the property. They feel that they can resell the property for $120,000 with probability of 0.15,

$100,000 with probability of 0.45, $80,000 with probability 0.25, and $60,000 with probability 0.15.

Compute and interpret the expected profit for reselling the property.

Let’s make a table to show what we know:

Selling Price ($) Profit ($) Probability Profit x probability

120,000 120,000 – 77,000 = 43,000 0.15 43,000 x 0.15 = 6,450

100,000 100,000 – 77,000 = 23,000 0.45 23,000 x 0.45 = 10,350

80,000 80,000 – 77,000 = 3,000 0.25 3,000 x 0.25 = 750

60,000 60,000 – 77,000 = -17,000 0.15 -17,000 x 0.15 = -2,550

Taking the sum of each product, my expected profit is .



In the previous example and related extensions, we discussed the general concept of discrete random

variables. Now we will look at a very specific type of discrete random variable known as the binomial

random variable which leads to the binomial probability distribution.

The binomial probability distribution is a discrete probability distribution that describes probabilities in

which there are two mutually exclusive (disjoint) outcomes, generally referred to as success and

failure.

Binomial Probability Distributions are created by binomial probability experiments, which are

experiments that:

Are performed a fixed number of times (trials)

Each trial is independent

Each trial has two disjoint outcomes: success and failure

The probability of success (and failure) is the same each trial

In a binomial probability experiment, the random variable X measures the number of success and is

therefore known as the binomial random variable.

Once you have established that a random variable is a binomial random variable, you can begin

thinking about computing the probabilities of the binomial experiment. The formula is:

P(x) = nCr px(1 – p)n - x

Let’s use a rolling number cube scenario to practice creating binomial probability distributions for the

discrete random variable X, rolling a 3. Remember X can be x = 0, 1, 2, 3, 4, or 5 based on rolling the

number cube 5 times. So we need to calculate the probabilities for rolling 0, 1, 2, 3, 4, or 5 threes out

of 5 total rolls.

x What we are actually calculating P(x)

0 5C0 x

0.402

1 5C1 x

0.402

2 5C2 x

0.161

3 5C3 x

0.032

4 5C4 x

0.003

5 5C5 x

0.0001



As you are analyzing and/or creating probability distributions, keep 2 things in mind:

3) All probabilities must be between 0 and 1

4) All probabilities must add up to 1.

Hopefully you are realizing how similar this process is to what we did earlier this semester with

frequency distributions. Just like before, we can also display these distributions in graph form,

especially histograms. Probability histograms are excellent visual aids for helping convey the likelihood

of particular (in this case, discrete) events. Below I have included a binomial probability histogram for

the discrete random variable rolling a 3.

Try This! Bob is an 82% free throw shooter. If we ask Bob to shoot 4 free throws, create a probability

distribution for the discrete random variable X, the number of free throws made. Use the probability

distribution to create a probability histogram for X.

x What we are actually calculating P(x)

0 4C0 x 0.001

1 4C1 x 0.019

2 4C2 x 0.131

3 4C3 x 0.397

4 4C4 x 0.452



Let’s look at the mean of a binomial random variable, the formula is In words, we simply

multiply the number of trials by the probability of success. Let’s practice using this formula using the

examples we have used above.

Example 1: Find the mean of the binomial random variable X, the number of 3’s rolled on a standard

number cube if the cube is rolled n = 5 times. Note, the probability of rolling a 3 is

.

So,

.

As most things go in statistics, how we interpret the results is very important. The way I interpret my

result is, If I roll a fair number cube 5 times I expect to get a three 0.83 times. This of course is a

theoretical value since it is based on purely what we expect to happen. I could always decide to run an

experiment to get an empirical value. Below is a summary of the results of running an experiment 20

times, the values represents how many times 3 appeared within each trial.

2 2 1 2 0 2 0 0 2 0

1 1 1 1 0 0 1 0 0 2

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0 1 2 3 4

Pro

bab

ility

Free Throws Made

Shooting Free Throws



If we take the mean of these values we will have , the mean number of 3’s rolled out of 5.

Notice how close is to what we calculated for . The more times we repeat the experiment, the

closer gets to .

Let’s practice finding the mean of the discrete random variable using the free throw example.

So if Bob shoots 4 free throws, we expect him to make 3.28 of them.

Of course you cannot make .28 free throws, that is why it is important to remember that this is

theoretical!

Below are the results of Bob shooting 50 sets of 4 free throws, calculate .

Below are the results of Bob shooting 100 sets of 4 free throws, calculate .

x P(x)

0 0.001

1 0.019

2 0.131

3 0.397

4 0.452

4 4 4 3 2

4 4 2 3 3

3 4 2 3 3

2 3 3 3 3

3 3 1 3 4

3 4 4 2 3

4 2 3 4 3

4 3 4 2 4

3 3 3 3 3

3 4 2 2 2

3 2 4 3 3 2 3 4 2 2

4 3 3 3 4 4 3 4 4 4

4 4 3 3 4 2 3 4 3 4

4 2 4 4 3 4 4 4 3 3



Notice when I increased the number of trials from 50 to 100, went from 3.06 to 3.29.

Recall, . Again, as n increase gets closer to .

When discussing binomial probability distributions, standard deviation can also be easily calculated.

Using the formula we can determine the spread of a binomial random variable.

Just to practice using the formula, the standard deviation of X in the example regarding rolling a

number cube would result in the following:

And in the free throw problem:

4 4 4 3 2 3 4 2 3 4

3 3 3 3 3 2 4 1 3 4

3 3 3 3 4 4 2 4 4 3

4 4 4 2 4 2 3 3 4 3

4 3 4 4 4 3 4 3 2 3

4 4 3 3 4 3 4 2 3 4

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Theoretical vs. Empirical Probabilities

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