>> Hey, everyone.
Thanks for joining us in a discussion
of binomial distributions.
As you can see, the first objective is to determine
if we've got an experiment that is a binomial experiment.
First, for an experiment to be considered binomial,
four things must hold.
You see the four things there listed.
The experiment that is performed needs to be
for a fixed number of trials.
Secondly, the trials need to be independent.
Third, there are only two mutually exclusive outcomes
for each trial.
This is the biggest piece of evidence in my opinion in terms
of being able to recognize, and of course,
we'll discuss this in the examples.
And finally, the probability of success needs to be the same
for each trial where we let P denote the probability
of success.
So, let's go ahead into our first example here
where they're asking us
to determine whether the following experiments are
binomial experiments and we have to explain the reasoning.
Taking a look at part A then, according to a recent study,
33% of Americans 23 years or older have been arrested.
A random sample of 500 Americans 23 years
or older are asked whether or not they have been arrested.
Remember we're trying
to determine whether this is a binomial experiment,
so first just to make it a bit clearer, we're going to go ahead
and state that this is a binomial experiment.
But we're going to explain why, of course.
It's pulling in the four criteria.
The first one, of course, is that the experiment is performed
for a fixed number of trials where they told us
that we have a random sample of 500 Americans.
Next, we need to recognize that the trials are independent.
That is, whether or not a randomly selected person has
been arrested does not affect whether
or not another randomly selected person has been arrested.
Therefore, the trials are independent Third,
and the pivotal one, there are only two outcomes,
either they have been arrested or not.
There's no other outcome, so we have the situation
that is only two outcomes.
And finally, the probability of success, being arrested this is,
is the same for each trial or P is 33%, or 0.33.
Let's jump over to part B, then, where we're told Mary is
at the fair playing pop the balloon with six darts.
There are twenty balloons total, fifteen which is say
"lose" and five which "win".
Now, this particular example is not a binomial experiment
since the trials are dependent and the probability of success
of winning changes with each trial.
And what we mean is when we pop the first balloon,
now the total number of balloons goes from twenty to nineteen,
and whether we win or lose
on that first one changes the probability regardless.
To explain a little more for some elaboration, we can set it
up like this where we're saying, okay,
what's the probability of winning?
Well, the probability of winning says that we have five balloons
that say win out of twenty total balloons.
Therefore, the probability that we will win
on the first dart would be a five out of twenty.
Now, if we went on the first dart and that means
for the second dart, the probability of we win changes
because now there are only nineteen balloons available
and with only nineteen balloons available we have only now four
of them that would allow us to win since we already first one
of the balloons that was a win.
Now, even if the first one was a lose,
it would still change the probabilities
because we would go from twenty to nineteen balloons.
So, we have dependent trials.
Let's now look at part C where Bob flips a coin
until the coin lands on heads.
Now, we have the idea that he's looking for heads.
There is the idea that is only two possible outcomes
where we have heads or tails.
However, the key piece here,
this is not a binomial experiment
since there is not a fixed number of trials.
They use the keyword until, meaning he's going to keep
on flipping until he gets a heads, so he may need one trial
or two trials or three trials until the coin lands on heads.
Pause the video and try these problems.
That takes us over to our second objective
where we will be calculating probabilities
for a binomial distribution.
This can be a bit tough based on the calculator input.
We have a table here to help us calculate probabilities
for these binomial distributions.
It's all based on the phrase that is be said to us,
for example, when the phrase is "exactly" or "equals" or "is",
what's happening is they are telling us
to find the probability of a specific outcome
with our math symbol being P of X being equal to some value X.
In the calculator, this is where we use the binom PDF function.
This is very specific, again, with a specific result.
As we continue the comparison, you see the other options
where they are set up with both the phrase
and then the corresponding math symbols.
You do have the option of when they tell us between A
and B inclusive there, we have the option
of adding these together with the binom PDF function.
Or we can use a combination of the binom CDF
where we find the probability of something happening and so track
that from the probability of outcomes that are not
in the event we're discussing.
Of course, we will refer back to this as we use our examples,
so let's go ahead and jump into those.
Let's first note, to get to either the binom PDF
or binom CDF functions in your calculator,
you press the second button, and then the vars [phonetic] button
and scroll up until you find either the binom PDF
or the binom CDF.
With our first example, we're told a study was done
which stated that 41% of Americans only have a cell phone
in their house, no landline that is.
What is the probability that in a random sample
of fifty American households
that exactly twenty only have a cell phone?
The first keyword that we're told is that the probability
that we're looking
for is exactly twenty only have a cell phone.
Translating this to our symbols, have a little bit
of language here, we're looking for the probability
that exactly twenty, which would be the probability that X,
the variable we're looking for, would be twenty.
Or X represents the number of households
that have only a cell phone.
Now, referring to the previous table,
remember since we have the value that is exactly twenty,
this tells us to use the binom PDF function.
Now, with the binom PDF function,
remember when you did input N, P,
and then X with commas separating those values,
which means we have to address each of those
from the given information in the problem.
N being the sample and they tell us up here
that we've got a random sample of fifty American households.
So, that means that N is equal to our fifty.
Next, we're looking for a P representing the probability
of success.
The study stated that 41% of Americans have only a cell phone
in their house, therefore the 41% represents the probability
of finding that success.
We, of course, write this as a decimal, as 0.41.
And lastly, X would be the number we're looking
for for the number of successes,
which we previously stated was twenty because they tell us
that we want exactly twenty in our problem.
If we input this into a TI-83, our screen ends up looking
like this with the binom PDF function.
On a TI-84 screen, it will end up looking like this.
If you have something called the stat wizard mode on,
or you can turn that off using the mode function,
with each calculator we should obtain approximately the same
result, getting a probability of 0.11.
And you're all set.
On to the next example were told according to a recent article,
38% of busses in Chicago arrive on time.
A random sample of thirty Chicago busses is taken.
In a random sample of thirty Chicago busses,
what is the probability that less than 10 arrive on time?
Which means, we're looking for the probability of less than 10.
Referring back to the table, we can translate this
as a math symbol and to X being less than 10.
In order to lead that to the correct input, now depending
on your approach or your instructor's approach,
we are able to put this directly to the binom CDF function,
referring to the table where we have strictly less than 10.
That is one route we can take is going to that table
where we say we've got a binom CDF now,
making sure to create some emphasis
on the fact that it's CDF.
And since it's strictly less than 10, that table tells us
to do N, P, and then more specifically this is an X minus
1, because what we are doing is we are excluding the value
of ten itself, which changes that to be a binom CDF.
And referring to the problem said
that we had thirty Chicago busses.
Therefore, we've got a value of 30 for N. The problem said
that 38% of busses arrive on time.
Therefore, our 38% represents the probability
of success, or P, being at 0.38.
And finally, making sure to be careful about it,
X minus 1 over X was set to be less than 10.
Therefore, we convert that to be a 10 minus 1, giving us 9.
Now, another way to think of this though is
to translate the original statement.
It all depends on your preference.
A lot of instructors will say instead of thinking of this
as P being less than 10 we can convert that statement.
We know that since we are working
with a discrete variable, if X is less than 10,
we can call that X being less than or equal to 9.
If we've got less than or equal to 9,
according to that table this would be equal to a binom CDF.
And on our table we see that it says N, P, and then just X. So,
the idea is whatever you're comfortable with.
We can think of it as less than 10.
If we think of it as less than 10, then we use X minus 1.
If we think of it as X less than or equal to 9 by converting
that to a less than or equal to, we've included the 9,
which allows us to just use X as our key number,
which means we end up with a binom CDF.
Our N is still 30, 0.38 is still P, and now our value for X is 9.
The result should be the same for each of them,
giving us a probability of 0.24, which means the probability
that less than 10 of the 30 Chicago busses will arrive
on time is about 24%.
Over to part B in our example then.
In a random sample of thirty Chicago busses,
what is the probability that exactly 17 arrive on time,
which means again we're referring to a keyword
that is saying exactly 17.
When they tell us exactly 17, that means we are trying
to find the probability that X,
the variable representing the number of busses that arrive
on time being specifically 17.
We know that since it's a specific value like that,
we need to use a binom PDF function as opposed to a CDF,
where N is still 30, P is still a 0.38 from previous,
and now X being the value 17 that we're looking for.
And put this in our calculator and we got ourselves a 0.02,
that is the probability that exactly 17
of the 30 Chicago busses will arrive on time is about 2%.
In part C of the example, we've got a random sample
of our 30 Chicago busses still.
What is the probability that at least 12 arrive on time?
The key phrase we're highlighting now is at least 12.
And we're looking for the probability of.
Make sure we're careful with this translation
into our math symbols because at least 12 includes the 12.
Therefore, we have the probability that X is greater
than or equal to the 12, meaning it's including that 12.
This again provides a couple opportunities
in order to translate this.
Referring back to the table, we can use the option
that is 1 minus the binom CDF function.
And as we see in the table, our input would be N, P,
and now an X minus 1 because it's including the 12.
The 1 minus of course means the complement of that.
If we use this, then I'm going 1 minus my binom CDF--
Where N is still 30, P was our 0.38, and X minus 1 needs
to be 12 minus our 1, getting 11.
Similar to the previous discussion, we can think of this
in terms of sample space and outcomes.
So, that means we can translate the probability
of X being greater than or equal to 12 into the complement of,
which would be our 1 minus, the probability
that X is less than or equal to 11.
If we do that, then that means we've got 1 minus a binom CDF--
Where N is still the number of trials, P, and since it's less
than or equal to that just stays X. And this leads us
to the same calculator input that we had
in the initial translation.
Either route we take leads us to the same result,
which would be a 0.48.
That is the probability that at least 12
of the 30 Chicago busses will arrive on time is 48%.
Finally, in part D, we have our random sample
of 30 Chicago busses still.
Question's asking us what is the probability that between 5
and 7 inclusive arrive on time?
Of course, that being our key phrase again, we're looking
for the probability that between 5 and 7 inclusive are on time.
We can translate that into saying that we are looking
for the probability between 5 and 7 inclusive,
or that is X is greater than or equal to 5 while also being less
than or equal two 7, which means remember we have two
possibilities for this approach as well.
We can piece this together as discrete inputs.
If we're piecing this together then
that means this would be equal to a binom PDF--
Where N is 30, P is still at 0.38,
and what we're doing is we're using the three options
that we're looking for, 5, 6, or 7, because remember,
they're discrete variables.
So, the first possible outcome we're looking for would be 5.
We add that to the second outcome we're looking for,
which would be 6 busses on time.
And we add that to the third option, that would be 7 busses
on time, leading us to a probability of a 0.07.
In your calculator you can put all three of those at once
or you can add them together.
The other route we can go conceptually would be to think
of this in terms of binom CDF function where we're looking
for the probability that we want.
And we're going to subtract that by the probability
that we do not want, which means our first input would be a binom
CDF now of 30 with P being 0.38
and that first value for X being 7.
What this would do would be the probability of 7
or less arriving on time.
But we don't want 7 or less.
We want between 5 and 7,
which means we subtract the probability that we don't want,
which tells us to do a binom CDF--
N still being 30, of course, 0.38 for our probability
of success, and now we want 7, 6, and 5.
Therefore, the first output
that we don't want would be 4 busses on time.
So, we're subtracting that out
from the probability we're looking for.
We can input that entire line all tighter in our calculator
if you want or piece by piece.
And we end up with the same result of a 0.07,
that is the probability that between 5 and 7 inclusive
of the 30 Chicago busses will arrive
on time is approximately 7%.
Pause the video and try these problems.
And thanks for joining us in a discussion
of binomial distributions.