- Hello.
I am Professor Rick Jerz,
and I welcome you back to my Week 3
reinforcement session of business analytics.
In this session, I plan to do several things.
I plan to start by giving you a
very easy probability example.
Yet, this example is difficult to calculate.
I will use this example to review
some of the probability concepts that we learned
in the concepts and Excel sessions.
Next, I would like to solve some of our problems
that we have seen in Moodle just to show you
how to apply what we have learned.
I will also provide you some logic
of how to approach and solve
some of the problems from this topic.
Finally, I will end with another simple
classic probability problem.
I always get excited and challenged
whenever I revisit this topic of probability.
Yep.
It is a tough topic.
But mastering some of the topics,
and being able to solve problems
is very rewarding and enjoyable.
Let's get started.
My objectives are to review a few administrative items,
to provide an example of where
calculating probability seems hard,
to solve some probability problems on Moodle,
to explore another interesting probability problem,
"Let's Make a Deal," and to say a few words
about our next session.
Relative to administrative items,
I cannot think of anything new for this week.
Let's explore some of the probability concepts
that we have learned by looking at
a very interesting problem called The Birthday Problem.
This problem I often do in a face-to-face classroom
when I have about 40 to 50 students.
Of course, I'm not with you in the classroom,
but I think I can use another group of 40 to 50 people
to illustrate the concept.
Here we go.
In a group of 40 to 50 people,
let's use our U.S. presidents.
I want to bet you 50 dollars that at least
two presidents will have a birthday
on the same day of the year.
For example, March 10.
Are you going to take my bet?
In a classroom, I of course would not know
the students' birth dates.
So let's assume that I do not know
the presidents' birth dates.
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
Remember that there are three ways of assessing probability.
If we were going to use the subjective approach
to this problem, we would just use our best judgment.
If we were going to take the empirical approach
to solving this problem, I would have to
perform this problem a number of times,
like maybe 50 times, and then see historically
how many of those 50 times two people
do have the same birthday.
Using the classical approach,
we would have to calculate the probability.
The classical approach is the most accurate,
but the challenge is how do we calculate the probability
in this particular problem?
Our experiment involves whether or not
people have the same birthday.
The various outcomes are that
no two people have the same birthday,
that two people have the same birthday,
three people have the same birthday,
four, five, six, up to, let's say our 50 people
involved in the experiment.
The event of interest is that two or more people
have the same birthday.
It would be great if we could calculate
what this probability is and show it as
a relative frequency diagram.
All we would need are two bars.
Yes, there are two or more people with the same birthday,
or no, there are not.
But at this point we don't have this diagram.
We still need to do the calculations.
This problem does give us a chance
to review some probability rules and concepts.
Remember that probability is a measure
of the likelihood that an event in the future will happen.
It is a value between zero and one.
The sum of all probabilities of
all possible outcomes equals one.
The outcomes must be mutually exclusive.
What this means is that we can't have
both five and six people with a birthday on the same day.
It's either five or six, for example.
Applying some of these concepts and rules,
our birthday problem probability distribution
might look something like this.
The bars shown in red represent our
probability of interest, that two or more people
have a birthday on the same date.
The typical way that some people approach this problem
is to start counting all of the various outcomes,
that two people have a birthday on January 1st,
that two people have a birthday
on January 2nd, and so forth.
Then, three people on January 1st,
three people on January 2nd, etc.
But you might already see that this gets pretty complex,
and it really does.
Well, there is another way that we might
calculate this probability.
Some would call this a trick,
but I would rather refer to it as
a clever way of using the complement rule.
Remember that the complement rule
says that the probability of an event of interest
is equal to one minus the probability of
the event not happening.
Let me illustrate this complement rule with two people.
Our question is, what is the probability
that they both have a birthday on the same day?
Let's assume that the first person
has a birthday on February 2nd.
We can calculate that there are 365 out of 366 days,
including leap year, that are not February 2nd.
So, in words, the probability that these two people
do have the same birthday is equivalent to
one minus the probability that they do not.
The probability is very small, .0027.
We can extend this idea to the third person.
The probability of the third person
not having a birthday on either the first
or the second person's day is 364 divided by 366,
meaning that there are 364 days
that are not the same as the first two people's birthday.
Each time we look at another person,
we have another event, and the probability of
this next event is conditioned by the previous events.
For each person, we have one less non-matching day.
We can use the rule of multiplication
with conditional probability to calculate the probability
of two or three or four people not having the same birthday.
Subtracting this from one gives us the probability
that they do have the same birthday.
Well, we still have a lot of calculations to make,
but I have constructed this Excel model
that does these calculations.
The first column represents the number
of people that we have.
The second column represents that this person's birthday
is not the same as the others before it.
The third column applies the rule of multiplication.
It calculates the product of all
of these previous probabilities,
and the last column subtracts the probability
that they are not the same from one,
and this gives us the probability that we do have a match.
As we scroll down this spreadsheet,
you will see that the probability of two students
having the same birthday increases
as the number of students increases.
Of course, if we had a group of 500 people,
certainly two people would have a birthday on the same day.
Now, let's return to my bet.
I bet you that of all of our presidents,
at least two have a birthday on the same day.
I have given you a link to a Wikipedia list
of all of the presidents and their birthday,
and you can easily sort this list by their birthday.
There have been 44 presidents, and if we look at
our Excel model, we would see that there is a 93.2%
probability that two have the same birthday.
You should notice that both Polk and Harding
had their birthday on the exact same day, November 2nd.
So, I win this bet.
Give me your 50 dollars.
This works for any group of 50 people.
You will win this bet more often than you will lose.
This represents the power of calculating probabilities.
If you know how to calculate probabilities,
you can win more often than you lose.
Let me now turn my attention to
helping you solve some of the problems in Moodle.
This chapter is basically divided into two parts.
The first part involves probability concepts
and the second part involves counting.
If a question that you see in Moodle
asks what is the probability,
you are into the first half
of the chapter involving concepts.
Maybe the rule of addition, the rule of multiplication,
or the complement rule.
Yes, you still have to decide what rule
you need to apply, but most likely,
if the question is asking about one event of interest,
you are into the rule of addition.
If it appears that the question is
asking about sequential events,
you are into the rule of multiplication.
If the question in Moodle asks,
how many, you are into the second half of the chapter,
and then you have to decide if the question involves
an arrangement, permutation, or combination.
For the second half of the chapter,
my Excel model can be used to help solve these problems.
On Moodle, I have given you a link to some of these
practice problems for this reinforcement session.
I am going to do a couple of these for you right now.
A developer of a new subdivision wants to build
homes that are all different.
There are three different interior plans
that can be combined with any of
five different home exteriors.
How many different homes can be built?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
When we read the question,
we are being asked for, how many?
So we are into the second half of the chapter,
arrangements, permutations, or combinations.
Since repetitions are allowed,
we are into an arrangement,
and all that we have to do is multiply three by five.
Yes, you could use my Excel model for this,
but the calculation is so simple I can do it in my head.
15.
Let's now do this problem.
The first card selected from a
standard 52-card deck was a king.
If it is not returned to the deck,
what is the probability that king will be drawn
on the second selection?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
When you read the question, it's asking for,
what is the probability?
So we are into the first half of the chapter.
We appear to have two events.
We draw a king, and then another king.
So it looks like we're into the law of multiplication
with conditional probability, since the king
is not returned to the deck.
But the question isn't asking for
the likelihood of a king followed by a king.
It's just asking for the probability
on the second event, so all we have
is the one event to worry about.
It is conditioned.
There are three kings left out of 51 cards.
So the probability is 3/51.
Let's do this problem.
A firm offers routine physical examinations
as part of a health service program for its employees.
The exams show that eight percent of the employees
needed corrective shoes, 15% needed major dental work,
and three percent needed both corrective shoes
and major dental work.
What is the probability that an employee
selected at random will need either
corrective shoes or major dental work?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
The first thing to notice is that the question
is asking, what is the probability?
So we are into the first half of the chapter.
Something about probability concepts.
I do not see any sequential events
represented in this question, so I think
that we are in the rule of addition.
We do have corrective shoes, or major dental work,
and if you didn't know anything about probability,
you would maybe just want to add eight and 15
and come up with 23%.
However, it does state that three percent
need both corrective shoes and major dental work,
so what we have here is the rule of addition,
with joint probability.
When we have joint probability,
it has to be subtracted from the total.
So in this case we have 8 plus 15, minus three, or 20%.
Let's do this problem.
The numbers zero through nine
are used in code groups of four
to identify items of clothing.
Code 1083 might identify a blue blouse, size medium.
The code group 2031 might identify a pair of pants,
size 18 and so on.
Repetitions of numbers are not permitted.
In other words, the same number cannot be used
more than once in a total sequence.
As examples, 2256, 2562, or 5559 would not be permitted.
How many different code groups can be designed?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
Okay.
When we read the question, it's asking how many?
So we are into the second half of the chapter.
It's either an arrangement, permutation, or combination.
We are being told that repetitions are not permitted.
This means that it is not an arrangement.
We're left with either permutations or combinations.
We cannot use the same digit more than once,
bu we can have different orders.
For example, we can have 1234, or 4321.
We can use my Excel model to solve this problem.
We have 10 different digits, zero through nine.
Remember the zero.
We are selecting four from the 10.
When we look at the definitions of permutations
and combinations, we do see that repetitions
are not allowed, but the sub-selection order
is important for permutations.
The sub-selection is this group of four,
and the order is important.
1234 is different than 4321.
Therefore, this is a permutation,
which gives us 5,040 ways.
Let's do one more problem.
A tire manufacturer advertises, the median life of our new
all-season radial tire is 50,000 miles.
An immediate adjustment will be made on any tire
that does not last at least 50,000 miles.
You purchase four of these tires.
What is the probability that all four tires
will wear out before traveling 50,000 miles?
I encourage you to pause this video,
and try to solve this problem on your own,
before seeing how I solve this problem.
Again, we see the question being,
what is the probability, so we are into
the first half of the chapter.
We do get one tire, then we get another one,
then another, and then another.
So this does have sequential events in it,
but I don't see any conditional probability.
The probability of any individual tire
lasting more than 50,000 miles is 50%,
because 50,000 is the median,
and the median is in the middle.
So we have four events for the four tires,
with a probability of 50% that they would wear out
before traveling 50,000 miles.
We have to multiply these four probabilities together,
because we are into the rule of multiplication.
Remember that 50% is equal to one half.
What we end up with is one half, times one half,
times one half, times one half, which is equal to 1/16.
Okay.
You can see the general approach to solving these problems.
The key in this chapter is to make sure
to ask me for help when you don't understand
how to solve one of these problems.
I'd like to turn my attention to
one more interesting probability problem.
It is sometimes the "Let's Make a Deal" problem.
Some of you might recall that there was
a game show on TV called Let's Make a Deal.
The host was Monty Hall.
Here's how it worked.
Monty Hall would invite somebody from the audience
to come up to the main stage.
The main stage had three doors, numbered one, two and three.
Monty Hall would explain to everyone
that behind one of these three doors was a remarkable gift,
oh, maybe something like a pink Cadillac.
Behind the other two doors were basically duds.
Maybe a broom, and maybe a pair of shoes.
The doors, of course, were closed,
and the contestant had to pick one of three doors.
This got kind of wild.
People in the audience screaming,
"Door number three! Door number two! Door number one!"
Well, let's assume that the contestant
would pick door number one.
The next thing that Monty Hall would do
is reveal one of the other two doors.
Let's say, in this case, door number three.
He would open up door number three
and show that, oh, door number three
has the pair of shoes.
What this meant was that the grand prize
was either behind door number one or door number two.
The next thing that Monty Hall would do
is ask the contestant if
they would like to switch their choice.
The contestant had the option of staying
with their original pick, door number one,
or switching to door number two.
The audience would go crazy and shout,
"Switch! Switch! Switch! Stay! Stay! Stay!"
The question that I have for you is,
what would you do?
Would you switch or stay?
Is there really any difference?
I encourage you to pause this video
and try to solve this problem on your own
before seeing how I solve this problem.
This Monty Hall Let's Make a Deal problem
became a classic probability problem.
Many people would say it made no difference.
When the contestant is down to two doors,
there is a 50-50 chance that either door
would contain the grand prize.
The solution to this problem got written about
and debated in professional journals.
The answer is that you should always switch.
Let's me explain why.
When the game starts, you have three doors,
and there is one-third probability
that each of the doors would have the prize.
When you pick a door, like door number one,
there is a one-third probability
that the prize is behind door number one,
and a two-thirds probability that it isn't.
Think about a line or a wall between your door,
door number one, and the other two doors.
On your side, you have one-third probability of
getting it right, and on the other side of the line,
or the wall, there's a two-thirds probability
that the prize is there.
When Monty Hall eliminates one of those two doors,
it doesn't change the probabilities that existed
at the beginning of the game.
The one remaining door on the other side of the line
still contains two-thirds of the probability
of being correct.
This is why the contestant should always switch.
Well, if you do not believe this yourself,
why don't you play this game with family members
and start tallying up whether switching or not switching
produces more wins.
This game show went off the air,
and I'm not sure if it was due to the fact
that knowing what the contestant should always do
caused this game show to flop.
I think that it does illustrate
that knowing how to calculate probabilities
produces some interesting results.
Before I leave you, let me mention that
in our next session, we're going to continue
talking about probabilities, probability distributions,
and we're going to look at an interesting
discrete probability distribution
called the Binomial Probability Distribution.
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.