("Are You Going With Me?" by Pat Metheny Group)
- Hello, I am professor Rick Jerz
and I welcome you back to my week four
Reinforcement session of Business Analytics.
In this session, I plan to do several things.
I plan to review just a few things
from this week's concepts and Excel videos.
Then we will look at one problem in depth
and tie some of these binomial probability concepts
back to some decision tree
and probability concepts.
Last, I will solve few binomial distribution problems
that you might see in Moodle.
Let's get started.
My objectives for this review session are
to review a few administrative items,
talk about using PivotTable to tally Ordinal data,
solve some binomial probability problems
using our Excel model
and to say a few words about next week.
Administratively, I don't need to say much
but I do want to say a few words
about the difficulty of these chapters.
In many courses, as the chapter number increases,
the level of difficulty also increases.
In general, this is true
of this particular course in Business Analytics.
However, I think you will find
that my Excel models make each successive chapter
be relatively easy.
So, chapter five is probably harder
than chapter six, seven, eight,
and so forth.
I'd like to say that our hardest chapter,
chapter five on probability,
is now behind us.
You might recall that a couple weeks ago,
I gave you a bonus challenge.
The challenge was to see
if you could use PivotTable
to properly order Ordinal Data.
I'm not going to show you some of the solutions
in this Reinforcement video,
but I will be providing you an administrative video
that shows some of the interesting solutions
to this problem.
I'd like to review the concept
of counting just one more time
using this rug problem.
This problem was presented in Moodle.
A rug manufacturer has decided to use
seven compatible colors in her rugs.
However, in weaving a rug,
only five spindles can be used.
In advertising, the rug manufacturer
wants to indicate the number
of different color groupings for sale.
How many color groupings using the seven colors
taken five at a time are there?
This assumes that five different colors
will go into each rug,
in other words,
there are no repetitions of color.
Visually, I like to think
of this rug as being round,
with five different rings, kind of like a target.
Let's explore this question
using three different assumptions.
The first assumption is that we can have
repetitions of colors.
If this was the case, in advertising,
how many different rugs would we
be able to say can be produced?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
This is an example of an arrangement,
for the first part of the rug,
we have seven colors that we can use.
For the second part, we also have seven.
The third seven, the fourth seven
and the fifth seven.
My Excel model does the arithmetic for us
and ends up with 16,807.
You can go off to any cell in Microsoft Excel
and do the calculation manually to prove
that this is the proper result.
Under this scenario, we could have a rug
with all of the same color.
For example, a completely blue rug
where all of the rings are blue.
Mathematically, this is what can happen.
It's still up to the business to decide
if they want to be selling rugs
with the same color.
Let's take another scenario.
Let's say that the business says,
we do not want to repeat any color in these rugs.
How many different rug styles can we advertise?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
The solution to this question depends upon
how we interpret five different colors.
If we have one rug made up of,
let's say, the primary colors from the outside in
red, blue, yellow, then white
and maybe pink.
Is that any different than reversing
this order from the outside in
going from pink, white, yellow, blue, red.
If it is, we would say that the order
does make the rug different.
In other words, order is important.
And this would be a Permutation.
We can manufacture 2,520
distinctly different rugs.
Another alternative way of asking this question
is saying that we cannot have repeats
of the same colors.
In other words,
our red, yellow, blue, white
and pink colors,
cannot be rearranged in any other order
to call the rug unique.
How many different rugs can we advertise
under this scenario?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
Well, this question really isn't all that difficult
because we only have one remaining method,
that's called a Combination.
And the answer to this question is 21.
So to answer the question,
how many different rugs might we advertise,
really depends upon the rules of selecting colors.
The only way to know is to
make sure to precisely define
what the question is being asked.
In this problem, as presented,
the question involves rug color groups
and therefore, I interpret this as not being allowed
to just mix up one group of colors
and calling it a different rug.
Therefore, I see this as a combination
and we have 21 different rugs
that we would be able to advertise.
Let's now switch our focus
to solving some Binomial Probability
distribution Problems.
An exam has three multiple choice questions
with five choices per question.
If somebody takes the exam
and just randomly guesses at the answer,
what is the likelihood
of getting exactly two questions correct?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
This problem can be solved in various ways
that we have learned to model problems.
Here is one approach using TreePlan.
Since for each question, there are five choices
with only one correct,
there is a one in five chance
of picking the right answer.
One in five is 20 percent
and that's what you see me modeling here in TreePlan.
You might notice that even though
our question seems relatively simple,
these decision tree models
get pretty complex, pretty quickly.
The first limb of this tree represents the first question.
The second limb, the second question
and the third limb, the third question.
At the end of each limb,
I use the number one or zero to represent
whether or not this particular path down the tree
produces exactly two correct answers.
You might notice that there are
only three paths down this tree
that produce exactly two correct answers.
So TreePlan calculates that the overall probability
of getting exactly two questions correct is .096.
This is about 10 percent
and it's not very likely
but it could happen.
To the upper right of this tree diagram,
I have created a table
that calculates these various probabilities
of getting from zero to three questions correct.
This becomes another way
that we can calculate probabilities.
In this case, however, we are using
the tree diagram to help us identify
which limbs have how many questions correct.
We wanted to know exactly two correct answers
and the table shows that there are three limbs
that produce two correct answers.
Each of these three limbs have their probabilities of .2, .2
that's correct times .8.
If you click on the probability cell,
you will see how I am doing this calculation.
Below this table, I have created a
relative frequency diagram
for these four possible events
of getting correct answers.
Zero, one, two, or three.
To the right of this table,
I show another way
that we can calculate this probability
using the Binomial Probability Equation shown above.
Of course, as you would expect,
using the Binomial Probability Equation,
we get the exact same results
as doing these calculations manually.
And with all three of these techniques,
the probability of getting exactly
two questions correct is still .096.
Another way that we could've gotten to the answer
to this question would've been
by using tables in the back of the book.
Down below, I am showing you the table
that we would have to use to do it this way,
n is equal to three.
Since we have three questions, the probability
of success is .2
and we are interested in exactly two correct.
This table shows that this probability is .096.
Okay, well, I have a bias.
I think that the easiest way
to solve these Binomial Distribution Problems
is to use the Excel model that I have provided to you.
All that you have to do here
is enter the probability of success being .2,
we have three questions
and we are interested in exactly two successes.
My model, to no surprise, calculates
the probability to be .096.
So what I have shown you are four different ways
to calculate the probability for
Binomial Probability Distribution questions.
But I still think that my model is the easiest to use.
For example, what if we were interested
in getting exactly two or three
questions correct out of the three?
What is the probability of this happening?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
Using my model, all we have to do
is enter three as the highest
and we see that the probability is .1040.
How about the probability
of getting less than two questions correct?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
Again, using my model, I enter one as the highest.
Remember the question is less than two
and zero as the lowest.
The probability is .8960.
This is interesting, there's a rather high probability
of getting only none or one question correct
on this exam if you are just guessing.
Now let me take a more realistic exam.
Let's assume that we have 20 questions,
each question have four multiple choice answers
and we are interested in knowing the likelihood
of getting, let's say, a C or C+
which could be 75 percent.
Meaning, more than 15 questions correct.
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
Alright, for 20 questions,
think about how you would have to do this
if you were only using TreePlan.
This tree diagram would be humongous.
Gee, maybe I should give this to you
as a bonus exercise,
just to kill about three hours
of your precious time.
I don't think so.
We could create a table using
the Binomial Probability Equation
and calculate these probabilities.
Maybe this would not take too long to do.
How about using tables in the back of the book?
You would think that this would be pretty quick
but if you were to try to do this,
you would run into a problem using the tables
in the back of our textbook.
There is no table
for n equals 20
and even if our exam had fewer questions,
we are saying that we have four multiple choice answers,
one is correct, the probability
of a correct answer is .25
and if you look at all the binomial probability tables,
there is no pi for exactly .25.
We'd have to choose between .2 or .3
and maybe resolve this issue
by doing some linear interpolation
between these two probabilities.
This sounds like a lot of work.
Using my model is definitely the fastest.
I enter .25 as the probability of success
then total number of trials is 20 questions.
The lowest number of successes is 16
and the highest is 20
and I see the probability of success
as being pretty low,
like zero percent.
In other words, if I were to give you an exam
with these 20 questions,
you would be much better off studying for the exam
and trying to do your best,
instead of just randomly guessing at the correct answer.
So once again, we are seeing how modeling some
of these probabilities scenarios in Excel
is really a handy way of doing these calculations.
We very quickly can calculate probabilities.
Let's do this one more problem from the textbook.
During the second round
of the 1989 U.S. Open Golf Tournament,
four golfers scored a hole-in-one on the sixth hole.
The odds of a professional golfer
making a hole-in-one are estimated to be 3,708 to one,
so the probability is one over 3,709.
There were 155 golfers participating
in the second round that day.
Estimate the probability that four golfers
would score a hole-in-one on the sixth hole.
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
So, in this problem, we are looking
for exactly four golfers.
You might notice that this problem is number 67.
It's a high number problem
which means that the author thinks that it's pretty hard.
And realistically, it is pretty hard.
The reason is that most students who are using tables
in the back of the book can't do this problem.
There is no table
for n equals 155.
So the author is forcing the average student
to have to use the binomial equation.
And of course, the author thinks
that this is pretty hard to do.
For us, we can use my Excel model
and solve this problem pretty quickly.
For the probability of success,
I'm not going to do this calculation manually,
I'm going to do it right in Excel.
I can enter
equal one
divided by 3,709.
The total number of trials is 155
and we are curious to know exactly four.
So I enter four as the lowest
and four as the highest number of successes.
I get the exact answer that our author gets,
it is not very likely,
nearly zero percent.
Okay, this is enough for now.
I think that I have convinced you that Excel
and my model make doing
these calculations pretty simple.
This is why I say chapter six
is actually easier than chapter five.
Next week, we are going to shift our focus
from these discrete Probability Distributions
to Continuous Probability Distributions.
The concept that we will use will be very similar.
We will have to create a diagram
for Continuous Probability Distributions
and see
what area under the curve
is answering the event of interest.
We will be building our statistical knowledge
on what we have learned thus far.
Thanks for spending some of your time with me.
I hope you have enjoyed
what I have shared with you.
If you ever have any questions
or need my help, please let me know.