Understanding Binomial Distributions

We are jumping into chapter 5, which we will discuss some discrete probability distributions. So in this video, we're going to start off with binomial distributions. So the binomial distribution centers around what is called a Bernoulli process. And a Bernoulli process is an experiment consisting of repeated trials. And these trials here are called Bernoulli trials. And each of these trials will have two possible outcomes, and they'll be labeled success or failure as one example, but generally labeled as success or failure with replacement such that the probability of success remains constant. And so then there's four properties that each Bernoulli process must possess, and it's basically a bulleted list of that first sentence. So the first property it must possess is that the experiment consists of repeated ... trials. So you have to have at least two things happening. It has to be repeated over multiple times. Secondly, each trial results in an outcome that may be classified as a success or failure. And so really what that means is that you should only have two distinct outcomes for each trial. So that's the idea behind success or failure. And then there's a couple other ways that we can define our success or failure, and we'll get to that in a second. But in general, you have to make sure with a binomial process, and one way you can also remember this is two means by. And so binomial means two distinct outcomes for each trial. The probability of success denoted by P remains constant from trial to trial. And the repeated trials must be independent from one another. And so we've talked a lot about statistically independent. situations. And so we need to make sure whenever we're doing a binomial problem, that our trials are independent of one another. And so they can't rely on each other. And so that's why we have this with replacement up here that just relates to this independence idea down here. So if we were talking about a deck of cards, and we pulled out a heart and did not replace that heart, and then we wanted to pull another heart, that would be be dependent on the previous scenario because now we're down a heart in our deck of cards because we did not replace it. So with replacement, that helps with this independence idea down here. Then we have our binomial random variable, which is the number X of successes in N Bernoulli trials. So N is the number of trials we are doing and X is our number of successes. And so our binomial variable that we're going to look at. is the number of successes. And so something that's going to be really important here, and we're going to touch on a couple different times throughout example problems and things like that, is what we define as a success depends. Sometimes the success of the problem will actually be the contextual failure. So if we're talking about a scenario where somebody does or does not get sick. So there's two potential outcomes there. Either they stay healthy or they get sick. We might define the get sick as our success, just make our calculations a little bit easier. But in real life, getting sick is considered a failure when the alternative is being healthy. And so we would want to define the bad or not great scenario of getting sick as our success, just because that's going to make our calculations a little easier. So. Keep in mind that success doesn't always mean the good thing in the problem. It's just what we want to know more about. Then we have our binomial distribution. So this is the probability distribution for our discrete binomial random variable. And so by definition, we have this long definition here. A Bernoulli trial can result in a success with probability of p and a failure with probability q equal to 1 minus p. So that's an important formula here. Then the probability distribution of the binomial random variable X, the number of successes in n independent trials is this equation here. So this is our binomial equation. Notice that this is a little f of X. And so if you recall, when we were talking about little f versus versus big F, little f of X is going to be the exact value for that number of X. And so we'll do an example problem where we talk about. little F and big F in a second, but know that this is your exact probability. for that number of successes, that x value. And so we're going to write it as f of x is equal to the binomial of x with n trials and p probability is equal to this is our combination and cx. That's just another way to write it. And so I highly, highly, highly recommend that you take a look at your calculator. You might have to Google and figure this out. But see if you can find this in your calculator somewhere, because it's way easier to type this in than to type all of this in. And so at the very least, your calculator will be able to do this formula. But if you can find out how to do this in your calculator, it's going to save you a lot of time on classwork, exams, etc. And one less formula you have to remember. So we're going to have the combination of n and x. And then we're going to take our p value, our probability of success. raised to our x times our q, which is 1 minus p raised to n minus x. And then the mean and variance of the binomial distribution, these are mu is equal to np, and sigma squared is equal to npq. So I think I've mentioned a couple times, we always have to remember that sigma is equal to the square root of sigma squared. And that's just something you got to keep in your mind at this point. So for this, our sigma is going to be equal to the square root of NPQ. And that's an easy translation from this. All right. So some samples. Oh, sorry about that. Some sample scenarios to help understand what a binomial or Bernoulli process might look like. You could have a scenario where you're looking at defects versus non-defects. And so you have two potential outcomes, either it's good or it's not. You could have a customer does or does not buy a product. You could have a component survives. or dies during a specific test, if you're doing some type of destructive testing, you could look at that burn early process. You could have that a patient does or not recover from a rare blood disease. You could have water is or is not impure. And those are just a couple examples. But in general, you're looking for a situation where either it is or it isn't. And whenever you see something like that happening, that's almost exclusively going to be a binomial distribution. The only thing you'll want to double check for is that they are independent. So when you're doing testing, you're. probably going to test something and then it's going to move on and you're going to test the next thing. And the quality or the, yeah, I'm going to say the quality of the first test is not going to impact the quality of the product in the second test because it's the same test and whatever happened in the first product has nothing to do with the second product. All right. So let's jump into an example here. It says find the probability distribution for the number of cars sold with a convertible top. for the next three cars sold. The historical probability of selling a convertible is 60%. So this three here, this represents our N. So we're going to have N equal to three. And something I want you to really start to get in the practice of if you haven't already is setting up your variables before jumping into the problem and setting them aside and picking them out. So here I'm pulling out my N value. And I know that this is my N because this is how many trials I have. have. I am buying, I have the next three cars to look at. So I have the first trial of the first car. The second trial is the second car. And then the third car would be my third trial. So I have three trials and the historical probability, probability of selling a convertible is 60%. So that's going to be my probability of success. So P is equal to 0.6. And I also know that this is my probability of success because what I want to know more about. is selling with a convertible top. And so with a convertible is a 60% success rate. Whereas if it was asking me to sell cars without a convertible top, I would actually use the 40%, the one minus P as my success. So I want to make sure that this value matches what I want to know more about. So then I can also, now that I have my P, I can find my Q, which is equal to one minus P. So that's going to be equal to 0.4. And then for this problem, thinking contextually, My X values are going to range from 0, 1, 2, or 3. And also note that this whole chapter is about discrete probabilities. So we're going to have whole numbers for everything that we're doing. So with this problem, I know that I'm ranging from 0 to 3 because it's the next three cars sold. So I can either have 0 convertibles out of those three. I can have one convertible. I can have two or I can have three convertibles out of the next three cars. So that's where my X values come from. So now I want to find my probability distribution. So the probability distribution is asking for my F of X for all values of X. So this upside down A indicates all values. So I need to find my f of x across all values of x. So I'm going to set up a table to help me with that. I'm going to do my x across the top, and then I'm going to do a little f of x. And then I'm also going to do my big f of x. So if you recall, big f of x is your cumulative probability. And then my x values are 0 to 3. So I'm going to have 0, 1, 2, and 3. Okay, now my table set up. So this value here is going to be my f, my little f of zero, which is equal to, so I'm going to write my formula here first, little f of x is equal to b of x n comma p, which is equal to n x p raised to the x q raised to the n minus x. So when I have f of zero, this is going to be the binomial of 0 comma n is 3, p is 0.6. And this is equal to 3, 0, p, or no, I lied, that should be 0.6, 0.6 raised to x. So that's 0. And then q is 0.4. So then I have 3 minus 0. is 3. And anything raised to the 0 is actually equal to 1. So I'm going to have 3C0, which is equal to 1, if I put that in my calculator, times 1 times 0.4 cubed, which is equal to 0.064. So now I can fill that into my table. So I can keep going with this formula, or I want to show you another way to think through this. So similar to the work that we've been doing, going back to our multiplication rule. So let's consider the scenario where we have X equal to one. So we have one convertible. This is going to be similar to our defect versus non-defect branching tree that we've been doing. where we have eight possible outcomes, we have three different pools of a binomial scenario. And so if we think back to that setup, when we have x equal to one, we can have the possible outcomes of convertible, non, non, non convertible, non, or non non convertible. And so for each of these in three separate ways. So we're going to have three times this. because each one's going to end up being the same. The probability of a convertible is 0.6. So this is our multiplication rule happening down here. 0.6 for convertible times the odds of a non-convertible is 0.4 times 0.4 for another non-convertible. And then we're multiplying it by three because we have three separate ways that this could happen. This is the same as taking each of these and adding them together. So 0.6 times 0. 0.4 times 0.4 plus 0.4, because now we have a non-convertible first, times 0.6 times 0.4 plus 0.4 times 0.4 times 0.6. So this and this are the same. And when you do that math, you end up getting the value 0.288. And would you believe it or not? If you do f of 1 equal to binomial of 1, 3, 0.6 equal to 3, 1, 0.61 times 0.4 squared, you get that same thing. Get 3 times 0.6 times 0.4 squared, which is equal to 0.288. So then I would come up here. I'm going to add a little bit. space to my table here. This is 0.288. So same thing with f of little two. If I were to plug in all those numbers, I end up getting 0.432. And for little f of three, I get 0.216. I'm going to copy those up here. You should take the time and run those calculations just to make sure that you know how to use your calculator for these problems, that you understand setting up those equations correctly. All right. So now I have my big F of X section left. I got to fill this in here. So my cumulative probability, if you recall, our big F of X is equal to the sum from X equal to zero. to my x of interest for little f of x. So it's everything from zero to that number. So when I'm doing big F of zero, that's from zero to zero. So I just need my zero value. So this is going to translate straight down. Now for big F of one, big F of one is equal to the sum from x equal to zero to one of f of x. So that's going to be f of zero. little f of 0 plus little f of 1. So I'm going to take 0.064 plus 0.288 and that gives me 0.352. Same similar thing for big F of 2. Big F of 2 is going to be 0.064. plus 0.288 plus 0.432. So that equals 0.784. And lastly, our final big F is the sum of all. And our sum of probabilities for every probability distribution should always add up to one. And so we can either do the math and double check, which is always a great idea. Or you can just write the number one. And so this is our final answer. This is our probability distribution for the number of cars sold with a convertible top for the next three cars sold. So now I'm going to show you how to use your binomial distribution table. So we've learned how to calculate our binomial distribution by hand using both the, um, to find our exact value, our little f of x. and also find our cumulative distribution, our big f of x. So we also have this really nifty tool called our binomial table. This is just a quick screenshot of what that looks like, just the first couple of numbers, but for this problem that's all we need. And so to use this, first of all, what it gives us is our big f of x, and we know that because of this summation right here. So this summation is telling us that the values in this table is everything from x equal to 0 to our r value. And so that's essentially this value here. This is equal to r in our table. And so for us, we are using n equal to 3 and p equal to 0.6. So we are going to be working with 0.6 and 3. So the entire time we're using this table, we're in this section down here. And so what that table gives us is our big S of X values from 0 to 3 at our p value of 0.6. And so you if you look, this is our big f of x table, we have point 0640.064. Then we have point 352.352. And then 784 and 1784 and one. So then we can also use our table here kind of in reverse to find our exact values. So if we wanted to find the exact value, of 1 occurring. So if we wanted little f of 1, if big F of 1 is equal to little f of 0 plus little f of 1, that means little f of 1 is equal to big F of 1 minus little f of 0. So that's some quick algebra for you. So what we would need to do there, and we also know that our little f of 0 is equal to big F of 0. So we can take our big F of 1 minus big F of 0. So that's going to be 352 minus 064, 0.352 minus 0.064. And that's going to give us 0.288. So if we go back up to our table, 0.288 is our little F of 1. So theoretically, using our binomial table, we can find all of these values here. Now, Complete disclosure, I'm going to ask you to do both. I need you to know how to find your little f and big F by hand and how to find both using the table. So that's why we talk about both in the video here. We'll play around with this in class. Be sure to bring any questions that you have in the class and we'll tackle them there.

And each of these trials will have two possible outcomes, and they'll be labeled success or failure as one example, but generally labeled as success or failure with replacement such that the probability of success remains constant. And so then there's four properties that each Bernoulli process must possess, and it's basically a bulleted list of that first sentence. So the first property it must possess is that the experiment consists of repeated ...

trials. So you have to have at least two things happening. It has to be repeated over multiple times. Secondly, each trial results in an outcome that may be classified as a success or failure.

And so really what that means is that you should only have two distinct outcomes for each trial. So that's the idea behind success or failure. And then there's a couple other ways that we can define our success or failure, and we'll get to that in a second.

But in general, you have to make sure with a binomial process, and one way you can also remember this is two means by. And so binomial means two distinct outcomes for each trial. The probability of success denoted by P remains constant from trial to trial. And the repeated trials must be independent from one another. And so we've talked a lot about statistically independent.

situations. And so we need to make sure whenever we're doing a binomial problem, that our trials are independent of one another. And so they can't rely on each other. And so that's why we have this with replacement up here that just relates to this independence idea down here. So if we were talking about a deck of cards, and we pulled out a heart and did not replace that heart, and then we wanted to pull another heart, that would be be dependent on the previous scenario because now we're down a heart in our deck of cards because we did not replace it.

So with replacement, that helps with this independence idea down here. Then we have our binomial random variable, which is the number X of successes in N Bernoulli trials. So N is the number of trials we are doing and X is our number of successes.

And so our binomial variable that we're going to look at. is the number of successes. And so something that's going to be really important here, and we're going to touch on a couple different times throughout example problems and things like that, is what we define as a success depends. Sometimes the success of the problem will actually be the contextual failure.

So if we're talking about a scenario where somebody does or does not get sick. So there's two potential outcomes there. Either they stay healthy or they get sick. We might define the get sick as our success, just make our calculations a little bit easier.

But in real life, getting sick is considered a failure when the alternative is being healthy. And so we would want to define the bad or not great scenario of getting sick as our success, just because that's going to make our calculations a little easier. So. Keep in mind that success doesn't always mean the good thing in the problem.

It's just what we want to know more about. Then we have our binomial distribution. So this is the probability distribution for our discrete binomial random variable. And so by definition, we have this long definition here.

A Bernoulli trial can result in a success with probability of p and a failure with probability q equal to 1 minus p. So that's an important formula here. Then the probability distribution of the binomial random variable X, the number of successes in n independent trials is this equation here.

So this is our binomial equation. Notice that this is a little f of X. And so if you recall, when we were talking about little f versus versus big F, little f of X is going to be the exact value for that number of X. And so we'll do an example problem where we talk about.

little F and big F in a second, but know that this is your exact probability. for that number of successes, that x value. And so we're going to write it as f of x is equal to the binomial of x with n trials and p probability is equal to this is our combination and cx. That's just another way to write it. And so I highly, highly, highly recommend that you take a look at your calculator.

You might have to Google and figure this out. But see if you can find this in your calculator somewhere, because it's way easier to type this in than to type all of this in. And so at the very least, your calculator will be able to do this formula. But if you can find out how to do this in your calculator, it's going to save you a lot of time on classwork, exams, etc. And one less formula you have to remember.

So we're going to have the combination of n and x. And then we're going to take our p value, our probability of success. raised to our x times our q, which is 1 minus p raised to n minus x. And then the mean and variance of the binomial distribution, these are mu is equal to np, and sigma squared is equal to npq.

So I think I've mentioned a couple times, we always have to remember that sigma is equal to the square root of sigma squared. And that's just something you got to keep in your mind at this point. So for this, our sigma is going to be equal to the square root of NPQ. And that's an easy translation from this. All right.

So some samples. Oh, sorry about that. Some sample scenarios to help understand what a binomial or Bernoulli process might look like.

You could have a scenario where you're looking at defects versus non-defects. And so you have two potential outcomes, either it's good or it's not. You could have a customer does or does not buy a product. You could have a component survives. or dies during a specific test, if you're doing some type of destructive testing, you could look at that burn early process.

You could have that a patient does or not recover from a rare blood disease. You could have water is or is not impure. And those are just a couple examples.

But in general, you're looking for a situation where either it is or it isn't. And whenever you see something like that happening, that's almost exclusively going to be a binomial distribution. The only thing you'll want to double check for is that they are independent. So when you're doing testing, you're. probably going to test something and then it's going to move on and you're going to test the next thing.

And the quality or the, yeah, I'm going to say the quality of the first test is not going to impact the quality of the product in the second test because it's the same test and whatever happened in the first product has nothing to do with the second product. All right. So let's jump into an example here.

It says find the probability distribution for the number of cars sold with a convertible top. for the next three cars sold. The historical probability of selling a convertible is 60%.

So this three here, this represents our N. So we're going to have N equal to three. And something I want you to really start to get in the practice of if you haven't already is setting up your variables before jumping into the problem and setting them aside and picking them out. So here I'm pulling out my N value. And I know that this is my N because this is how many trials I have.

have. I am buying, I have the next three cars to look at. So I have the first trial of the first car. The second trial is the second car.

And then the third car would be my third trial. So I have three trials and the historical probability, probability of selling a convertible is 60%. So that's going to be my probability of success.

So P is equal to 0.6. And I also know that this is my probability of success because what I want to know more about. is selling with a convertible top.

And so with a convertible is a 60% success rate. Whereas if it was asking me to sell cars without a convertible top, I would actually use the 40%, the one minus P as my success. So I want to make sure that this value matches what I want to know more about.

So then I can also, now that I have my P, I can find my Q, which is equal to one minus P. So that's going to be equal to 0.4. And then for this problem, thinking contextually, My X values are going to range from 0, 1, 2, or 3. And also note that this whole chapter is about discrete probabilities.

So we're going to have whole numbers for everything that we're doing. So with this problem, I know that I'm ranging from 0 to 3 because it's the next three cars sold. So I can either have 0 convertibles out of those three.

I can have one convertible. I can have two or I can have three convertibles out of the next three cars. So that's where my X values come from.

So now I want to find my probability distribution. So the probability distribution is asking for my F of X for all values of X. So this upside down A indicates all values.

So I need to find my f of x across all values of x. So I'm going to set up a table to help me with that. I'm going to do my x across the top, and then I'm going to do a little f of x.

And then I'm also going to do my big f of x. So if you recall, big f of x is your cumulative probability. And then my x values are 0 to 3. So I'm going to have 0, 1, 2, and 3. Okay, now my table set up. So this value here is going to be my f, my little f of zero, which is equal to, so I'm going to write my formula here first, little f of x is equal to b of x n comma p, which is equal to n x p raised to the x q raised to the n minus x. So when I have f of zero, this is going to be the binomial of 0 comma n is 3, p is 0.6.

And this is equal to 3, 0, p, or no, I lied, that should be 0.6, 0.6 raised to x. So that's 0. And then q is 0.4. So then I have 3 minus 0. is 3. And anything raised to the 0 is actually equal to 1. So I'm going to have 3C0, which is equal to 1, if I put that in my calculator, times 1 times 0.4 cubed, which is equal to 0.064.

So now I can fill that into my table. So I can keep going with this formula, or I want to show you another way to think through this. So similar to the work that we've been doing, going back to our multiplication rule.

So let's consider the scenario where we have X equal to one. So we have one convertible. This is going to be similar to our defect versus non-defect branching tree that we've been doing.

where we have eight possible outcomes, we have three different pools of a binomial scenario. And so if we think back to that setup, when we have x equal to one, we can have the possible outcomes of convertible, non, non, non convertible, non, or non non convertible. And so for each of these in three separate ways. So we're going to have three times this. because each one's going to end up being the same.

The probability of a convertible is 0.6. So this is our multiplication rule happening down here. 0.6 for convertible times the odds of a non-convertible is 0.4 times 0.4 for another non-convertible. And then we're multiplying it by three because we have three separate ways that this could happen. This is the same as taking each of these and adding them together.

So 0.6 times 0. 0.4 times 0.4 plus 0.4, because now we have a non-convertible first, times 0.6 times 0.4 plus 0.4 times 0.4 times 0.6. So this and this are the same. And when you do that math, you end up getting the value 0.288.

And would you believe it or not? If you do f of 1 equal to binomial of 1, 3, 0.6 equal to 3, 1, 0.61 times 0.4 squared, you get that same thing. Get 3 times 0.6 times 0.4 squared, which is equal to 0.288.

So then I would come up here. I'm going to add a little bit. space to my table here. This is 0.288. So same thing with f of little two.

If I were to plug in all those numbers, I end up getting 0.432. And for little f of three, I get 0.216. I'm going to copy those up here. You should take the time and run those calculations just to make sure that you know how to use your calculator for these problems, that you understand setting up those equations correctly.

All right. So now I have my big F of X section left. I got to fill this in here.

So my cumulative probability, if you recall, our big F of X is equal to the sum from X equal to zero. to my x of interest for little f of x. So it's everything from zero to that number.

So when I'm doing big F of zero, that's from zero to zero. So I just need my zero value. So this is going to translate straight down.

Now for big F of one, big F of one is equal to the sum from x equal to zero to one of f of x. So that's going to be f of zero. little f of 0 plus little f of 1. So I'm going to take 0.064 plus 0.288 and that gives me 0.352. Same similar thing for big F of 2. Big F of 2 is going to be 0.064. plus 0.288 plus 0.432.

So that equals 0.784. And lastly, our final big F is the sum of all. And our sum of probabilities for every probability distribution should always add up to one.

And so we can either do the math and double check, which is always a great idea. Or you can just write the number one. And so this is our final answer. This is our probability distribution for the number of cars sold with a convertible top for the next three cars sold. So now I'm going to show you how to use your binomial distribution table.

So we've learned how to calculate our binomial distribution by hand using both the, um, to find our exact value, our little f of x. and also find our cumulative distribution, our big f of x. So we also have this really nifty tool called our binomial table. This is just a quick screenshot of what that looks like, just the first couple of numbers, but for this problem that's all we need. And so to use this, first of all, what it gives us is our big f of x, and we know that because of this summation right here.

So this summation is telling us that the values in this table is everything from x equal to 0 to our r value. And so that's essentially this value here. This is equal to r in our table. And so for us, we are using n equal to 3 and p equal to 0.6. So we are going to be working with 0.6 and 3. So the entire time we're using this table, we're in this section down here.

And so what that table gives us is our big S of X values from 0 to 3 at our p value of 0.6. And so you if you look, this is our big f of x table, we have point 0640.064. Then we have point 352.352.

And then 784 and 1784 and one. So then we can also use our table here kind of in reverse to find our exact values. So if we wanted to find the exact value, of 1 occurring.

So if we wanted little f of 1, if big F of 1 is equal to little f of 0 plus little f of 1, that means little f of 1 is equal to big F of 1 minus little f of 0. So that's some quick algebra for you. So what we would need to do there, and we also know that our little f of 0 is equal to big F of 0. So we can take our big F of 1 minus big F of 0. So that's going to be 352 minus 064, 0.352 minus 0.064. And that's going to give us 0.288. So if we go back up to our table, 0.288 is our little F of 1. So theoretically, using our binomial table, we can find all of these values here.

Now, Complete disclosure, I'm going to ask you to do both. I need you to know how to find your little f and big F by hand and how to find both using the table. So that's why we talk about both in the video here. We'll play around with this in class. Be sure to bring any questions that you have in the class and we'll tackle them there.

Transcript for:Understanding Binomial Distributions

Transcript for:
Understanding Binomial Distributions