Binomial Distribution - Online Tutor, Practice Problems & Exam Prep

The Binomial Experiment

In recent videos, we've talked a lot about discrete random variables, numbers that can't be broken down any further, like the number of prizes in a raffle or the number of kids in a household. But there are many possible outcomes in those situations, like 0, 1, 2, 3, 4, 5, and so on and so forth. Now, a lot of times we will be looking at situations where there are only 2 possible outcomes, like heads or tails in a coin flip. So, one of the most common examples of a discrete random variable is the outcomes in what we call a binomial experiment. We'll talk a lot about probabilities, means, and standard deviations in a later video.

But in this video, I just want to introduce you to what it means for an experiment to be binomial. There are basically these four criteria that we're going to look for. So let's go ahead and take a look, and we'll do some examples together. Alright? So a binomial experiment, basically, the word binomial just means that there are 2.

These are events with 2 possible outcomes, either a success or a failure. Those don't necessarily mean something good or bad. So, for example, heads could be a success, and tails could be a failure, not necessarily good or bad. Alright? The variable little x that we'll be using in these problems is basically the number of successes.

So in other words, this is your discrete random variable. So what is the word experiment? So, basically, what is happening is you're just going to be doing a bunch of trials, a series of independent trials. So if I flip a coin 4 times, that's just 4 independent trials. The word independence we've actually seen before.

This just means that the outcome of one trial doesn't affect the next. If I flip a coin, it's going to land on heads or tails, then has no effect on what the next coin flip is going to be. Alright? Each trial has a probability of success. For example, the probability I land on heads, and the variable we use for that is the letter little p.

Each trial also has a probability of failure, and these successes and failures are not always going to be 50/50. The letter that we use for this is q. And, basically, we've seen the relationship between complements before where the probability of something not happening, if there are only 2 possible outcomes, is 1 minus the probability that it does. So, basically, what happens here is if failure is the opposite of success, then it means that the probability of failure is just 1 minus p. Usually, what's going to happen in these problems is they'll give you what the probability of success is, and you can always figure out what q is using this relationship.

Alright? So that's all the variables you need to know. Let's take a look at our example here. So we're going to take a look at these following experiments and figure out if they are binomial. And if they are, we're going to figure out the values of n, p, q, and little x.

Alright? So let's go ahead and get started with our first example here. You're going to flip a coin 4 times. You count the number of times the coin lands on heads, which is 3 times. Okay?

So is this binomial? There are basically 4 questions to ask here, which have to do with the information that we just talked about up here. So let's get started. The first thing you have to answer is, are there only 2 possible outcomes? Well, if you flip a coin, you either get heads or tails.

There's no other possible outcome. Right? So there are only 2 outcomes in this experiment. So are there a fixed number of trials? Are you doing this a set number of times, or are you flipping the coin until something happens?

Well, basically, what we're told here is that you're flipping the coin exactly 4 times, so that is a fixed number of trials. Are the trials independent, which means that the outcome of 1 affects the others? We've already talked about this. One coin flip doesn't affect the next one, so these are definitely independent. And is there an equal probability of success per trial?

Oh, in other words, if success is the number of heads that I get, then it's always 50/50 because they are independent. So usually what happens is that these last two questions are sort of tied together. Alright? So this is definitely a binomial experiment. Let's take a look at the values of n, p, q, and little x.

Number, n is basically just the number of trials. And so in this case, what happens is we're flipping the coin 4 times. So that is our n. Right? So n equals 4.

So what about p, the probability of success? Well, essentially, with a coin flip, it's 50/50, but we usually write these as decimals. So in other words, p is 0.5, and that just means that q is equal to 1 minus 0.5, which is also 0.5. Now don't be really careful here. Just because this coin flip is 50/50, doesn't mean that all binomial experiments are going to be 50/50 like this.

We're going to see other types of probabilities. And finally, our last variable is x, which is essentially the number of successes that we got. Well, the desired outcome that we wanted is we wanted heads, and we landed on that three times. So that is basically what little x is. Alright?

That's just the number of desired outcomes. So that is for this first example. Let's take a look at the second one over here. We're going to pull 4 marbles out of a bag of red and blue marbles, but we're not going to replace them. So we're not replacing

A basketball player normally has a 70% chance of making a free throw. The player shoots until finally making a basket, where $x$ is the number of shots they take. Is this a binomial experiment?

• A. Yes
• B. No, there are not 2 outcomes
• C. No, there are not a fixed # of trials
• D. No, the trials are not independent
• E. No, the probability of success is not the same for each trial

The Binomial Experiment Example 1

Back, everyone. Let's take a look at this example problem. We're going to be looking at rolling a 6-sided die. Let's take it from the top. We're going to roll a 6-sided die 10 times.

It's just a standard, you know, 6-sided die like this. You're going to record the number that you land on for each roll as the random variable. Is this a binomial experiment? So remember, there are basically 4 criteria questions we have to ask and answer. And let's go ahead and take it from the top.

Actually, we're going to skip this first one and come back to it later. So are there a fixed number of trials? You're going to roll this 6-sided die 10 times, so that is definitely a fixed number of trials. That condition definitely applies. Are the trials independent of another?

Does the outcome of one affect the rest? Well, if I roll the dice once and whatever that number lands on, that doesn't have any effect on the outcome of the next one. So there's definitely independence in the trials. And is there an equal chance of success? Well, however you're defining success, there's definitely an equal chance of success there.

So what about this 2 outcomes thing? Does this condition apply? Well, if you take a look at the wording of the question, you're going to record the number that you land on for each roll as the random variable. Remember, in a 6-sided die, there are 6 possible combinations or 6 possible outcomes, 1, 2, 3, 4, 5, or 6. So because you're recording the number that you land on, there are 6 possible outcomes in that experiment.

So because this criterion fails, what happens here is this is not a binomial experiment. So the answer for this specific question over here is b. There are not 2 possible outcomes. Alright? So, hopefully, that makes sense.

Now let's take a look at the next problem here. Using the above example, now you roll another 10 times, so you're basically conducting another fixed number of trials, and you choose the random variable. So this is the important part. You're choosing the random variable as the number of times you roll exactly a 2. So you're going to roll the dice 10 times, and you're going to count up the number of times you get a 2.

Is this a binomial experiment? Well, let's go through the same criteria again, and let's take a look. So now what happens is now that you've changed what you're defining as the random variable, as the number of times you roll exactly a 2, are there two possible outcomes? All of the other conditions apply here. You have a fixed number of trials, independence, and equal probability of success.

So are there 2 possible outcomes? What happens here is yes. Even though there are 6 possible numbers that you could get on the dice, because you're defining the random variable as the number of times you roll exactly a 2, there are 2 possible outcomes. Either you land on a 2 or you land on one of the other five numbers. Right?

So that would be a success. This would be a success here, and rolling the other five numbers would be a failure. There's still 2 possible outcomes even though those aren't the same. Alright? So all of these other conditions apply.

And because you've changed the wording of the random variable, this is a binomial experiment. Alright? So hopefully that makes sense. The rest of these are wrong. And let me, let's go ahead and take a look at the next one.

Thanks for watching.

The Binomial Experiment Example 2

Welcome back, everyone. So let's take a look at this example problem here. We have a pharmaceutical company that is developing a new vaccine. They claim the probability of gaining immunity from this new vaccine is 80%. Now the vaccine is given to 100 patients, and what happens is 83% or 83 of them gain immunity.

So we're going to figure out if this is a binomial experiment. And if it is, we're going to identify the values of npqandlittlex. Alright? Remember, the first part here is figuring out whether this is a binomial experiment, and just remember that these four conditions have to apply. Let's go ahead and take a look here.

Are there 2 possible outcomes? Well, you roll out a new vaccine, you give it to a person, and they either get immunity or they don't. There's no third possible outcome there. So there are definitely 2 possible outcomes. Are there a fixed number of trials?

Well, the vaccine was given to a set number of people, 100 patients, and then you're trying to count up the number of people who do gain immunity. So there's definitely a fixed number of trials here. Are the trials independent from one another? In other words, if I give the vaccine to one person, does that outcome affect the next person I give it to? And, no, it doesn't.

So in other words, these are definitely independent trials. And is there an equal probability of success per trial? Well, we're actually just told that the probability of gaining immunity, regardless of who you are as a person, is just a flat 80%. So there's definitely an equal chance of success. So this is definitely a binomial experiment.

Alright? So now because this is, we're going to figure out the values of n, p, q, and littlex. What is n? Remember, n is always just the number of trials that you're taking.

In this case, there are a fixed number of trials because the vaccine was given to 100 patients. So n = 100. What's the probability of success, which, remember, is just p?

Basically, what we're told here is that the probability of success, the desired outcome, is just that flat 80%. Normally, when we write probabilities, they're expressed as decimals, not percentages. So I'm just going to write this as 0.8.

Okay. What about q? Remember, q is the probability of failure. We're not told what it is explicitly in this problem, but remember, there's always a relationship between p and 1. If p is the probability of success, q is the probability of failure.

It's just 1 minus p. So now, the last but not least, what is littlex? Remember, littlex is your discrete random variable. It's the number of successes that you're looking for. And in this case, the number of successful outcomes was the number of people who gained immunity, which is 83. So you had 100 trials, but 83 of them actually were successful. That would be your littlex.

So that's all the variables in this binomial experiment. Let me know if you have any questions. Let's move on.

Calculating Probabilities in a Binomial Distribution

A lot of problems will give you a binomial experiment and ask you to calculate the probability of getting an exact number of successes. For example, you flip a coin 10 times, what's the probability you get exactly 4 heads? So you might think that we're going to have to use some complicated tables or maybe even list a bunch of outcomes to solve this. But what I'm going to do in this video is show you a step-by-step process of how to identify all the variables in your problem, and then we're just going to use one formula to calculate this. It seems a little scary at first, but it's actually going to involve a lot of things we've already talked about, like factorials and such.

Alright? I'm going to walk you through a step-by-step process. Let's just jump right into our example and see how this works. So, we have a bunch of red and blue marbles in a bag.

You're going to take one marble out and then replace it, meaning it's an independent trial, and then you're going to repeat this for a total of 4 times. The probability of pulling out a red marble is 80%. And what's the probability of getting exactly 3 red marbles?

So let's take a look at the first step over here. There's a lot of stuff going on, a lot of numbers. The first thing you want to do is identify what it means to be a trial, a success, and a failure, just kind of trying to understand your problem a little bit. Basically, what's happening is you put a bunch of marbles in a bag, and you're pulling out one of them and replacing it. So one trial is one marble.

So what is success? Well, basically, that's a desired outcome, and you're trying to get exactly 3 red marbles. So a success would just be me pulling out a red marble, and I'll just put r for this. That means a failure, not necessarily something bad, is just when I pull out some other thing over here. There's only 2 outcomes.

I only put either pull a red or a blue marble out. So failure is me pulling out a blue marble. Okay? So once we're done figuring out what our successes and failures are, now we're just going to go ahead and dig into the variables a bit more. There are basically 4 things we have to identify, npq, and little x.

Remember, n is the number of trials. So, essentially, how many marbles am I pulling out here? I'm pulling out 4 marbles and replacing them one at a time. So it's independent. One outcome doesn't affect the others.

So my n is equal to 4. Alright? What is the probability of success? Remember, that's little p. Basically, we're actually told that explicitly, and usually you will, it's 80%.

So in other words, there's an 80% likelihood that we'll get a success. So that's 0.8. Alright? So what about failures then? Remember, whenever you're given p, you can figure out q because it's just 1 minus your p.

Alright? So it's 1-0.8, which is 0.2. There's a 20% chance that I pull out a blue marble, and that's a failure. Okay? The last thing I have to identify is x, which is the desired number of successes.

What am I actually trying to find here? I'm trying to find the probability that I get exactly 3 red marbles. In other words, my little x is equal to 3. So one of the ways you might see this written out is p big x is equal to 3 like this, or you might actually just see this like p (3). Either way works.

P(X=x) = n ! x ! n − x ! ⋅ p x ⋅ q n − x

Alright? So, basically, this is what's going on here. I'm going to set up a bunch of probabilities. Essentially, what I'm trying to find here is I'm trying to find the probability of multiple events happening, like the probability that I pull out a red marble and then pull out another red marble and then pull out another another red marble. So I'm using the 'and' rule.

Remember, when these things are independent events, I basically just multiply their probabilities. So, in other words, the way I set this up is, what's the probability that I'd pull out a red marble, and then another red marble, and then another red marble, and then a blue marble. Alright? So I basically have 4 of these things that I want to multiply together. The problem is before I start multiplying numbers, this isn't the only way this experiment could have been successful.

I could have pulled out the blue marble first and then pulled out the red marbles, the 3 ones, and that would have also been a successful outcome. So this isn't the only way that I can get 3 red marbles. So what I have to do is after I'm done multiplying all these things, I have to multiply it by this term, which is essentially the number of ways that I can arrange 3 red marbles out of 4 complete trials. Alright? And so, basically, what happens here is this is just going to be my probability of getting a red marble, which, remember, is just 0.8.

And I'm going to raise this to the 3rd power. Then I'm going to multiply this by this is going to be the probability that I pull out a blue marble, 0.2, raised to the 1 power, and you'll see why this works in just a second. And I'm going to multiply this by the number of ways that I can arrange 3 red marbles out of 4 possible objects. So we've actually dealt with this before.

This is basically a combination. So you may see this notation over here, n c, or you may actually see something like this. But we've actually seen how to do this before. What, essentially, I'm trying to do is try to figure out how many times can I do 4 choose 3? And to do this, I'm just going to take n factorial, where n is the number of trials.

So in other words, that's 4 factorial divided by 3 factorial and then 4 − 3 factorial. Basically, really quickly, this just becomes a 1. And then what happens is

You take a 6-question quiz with True/False questions. What is the probability of getting all 6 questions correct by simply guessing?

• A. 0.015%
• B. 0.50%
• C. 5.6%
• D. 1.56%

National surveys indicate that 36% of people have been in a car accident in the last 5 years. If you randomly sample 10 people, how likely is that exactly 4 have had an accident in the last 5 years?

• A. 0.0767
• B. 0.242
• C. 0.00527
• D. 0.00129

Mean & Standard Deviation of Binomial Distribution

Hey, everyone. So in previous videos, when we were talking about discrete random variables, two quantities we were often interested in were the mean or the expected value and the standard deviation. Now you may also have to find those quantities when we're talking about binomial distributions. And whereas you could use these equations, oftentimes they involve summations and squares and things like that. So what I'm going to do in this video is I'm going to show you that instead of using these complicated formulas, we can actually find μ and σ, the mean and standard deviation of a binomial distribution using some shortcut equations that only just involve n and p.

Let's take a look at our example, and I'll show you how this works. It's pretty straightforward. So, we have a company that's launching a new product, and we're told that 60% of customers are expected to purchase it after they watch a demonstration. You randomly select 10 customers. How many would you expect to purchase this product?

Whenever you see those words "would you expect," they're asking for the expected value. And remember that in these types of discrete random variables, the expected value and the μ are synonymous. They mean the same exact thing. So basically, they're asking you to find the mean. So let's start out with that first.

When we did this for discrete random variables, we had to sum basically a bunch of probabilities times outcomes. For binomial distributions, it's much more straightforward. Basically, the equation is just n times p. That's all there is to it. So in this equation over here, we have n is equal to the number of trials.

In this case, we're talking to 10 customers, so that's our number of trials. And the probability of success, we're told, is 60%. 60% are expected to buy this thing. So, basically, we're just going to multiply these two numbers. So in other words, the μ or the expected number here is basically just going to be 10 times 0.6, which equals 6 customers.

Alright? So this makes sense if you think about this. Right? Remember the mean or the expected value? Basically, if I just grabbed a bunch of random samples of 10 people, if the probability of them purchasing the product is 60% and I grab 10 people, then the expected value that I should see is about 6 customers per sample of 10. And that's exactly what this says. Alright? So 6 customers is your mean. What about the standard deviation? So we're going to find the standard deviation and the variance of this distribution.

Alright? So what's going to happen here is basically the standard deviation is going to be a similar equation. We're going to take the square roots of n times p times little q. That's it. Alright?

So in other words, we're going to go over here. So our σ_x for this distribution is just going to be the square root √ (n × p × q). We've got 10 times 0.6. That's our probability of success, which means that our probability of failure with little q is going to be 0.4.

And we're just going to multiply all those numbers. What you should get over here is you should get something like 2.4. Actually, you'll get exactly 2.4. And then when you take the square root of that, you're going to get 1.55. So 1.55 customers is your standard deviation.

Okay. So that's the second question. That's the standard deviation. I'm going to mark that in blue. And then our last thing we have to find is just the variance.

So remember that the relationship between variance and standard deviation, once you know one, you always know the other. So the variance is just going to be σ². And basically, one way you can find this is you basically just take the square roots off of this formula over here. It's just n × p × q, which is actually just going to be this number over here in the square root. So it's basically just going to be 2.4.

Alright? So again, this is 2.4. Technically, this would be customers squared, but usually you don't put units to this. But this is your answer. This is the mean and this is the mean.

This is the standard deviation and the variance. Alright? That's all there is to it, folks. So let me know if you have any questions. Let's go ahead and take a look at some practice.-

Based on historical weather data in a certain city, about 62% of the days are cloudy. Find the mean, standard deviation, and variance for the number of cloudy days in a 30-day month.

• A. Mean = 18.6 days; Standard deviation = 2.66 days; Variance = $7.08days^2$
• B. Mean = 18.6 days; Standard deviation = 3.38 days; Variance = $11.4days^2$
• C. Mean = 11.4 days; Standard deviation = 2.66 days; Variance = $7.08days^2$
• D. Mean = 11.4 days; Standard deviation = 3.38 days; Variance = $11.4days^2$

Finding Multiple Probabilities in Binomial Distribution

So in the last couple of videos, we saw how to find the probability of an exact number of successes when we're given a binomial distribution. But in a lot of problems like the one we're going to work out down here, you'll be asked to calculate the probability of between two numbers of successes happening, or you may see other words, such as at least, greater than, less than, or something like that. And all that's happening here is that when you see these keywords, we're still just going to use this problem, this formula here, to solve these problems, but we're just going to have to add probabilities. Alright? So there's nothing new here.

We're basically just going to be adding a bunch of these formulas stacked on top of each other. Let's go ahead and walk through this example, and I'll show you how this works. Alright? So we have a marketing company that wants to analyze how many customers will open up a promotional email. So they send it to 10 customers.

In other words, that's the number of trials. Right? \( N = 10 \). So that's over here. Based on previous ad campaigns they've done or marketing campaigns, they find that 40% of the people typically open up the email.

However, that's the desired outcome, so that's the probability of success. That's \( p \), which means that, by the way, \( q \) is equal to the probability of failure, which is 0.60. Okay? So we have \( np \) and \( q \). Let's take a look at the first problem here.

The first problem says the probability that between 0 and 2 customers open up the email. This is not the probability that exactly 0 or exactly 2. It's between 2 and 2. And basically, all that means here is you're just going to add up where \( p \) is equal to the smaller number, which is 0, and you're just going to add all the way up until you get to whatever the bigger number is, so 2. So basically, between 0 and 2 means, if I sort of expand this, it's the probability that \( x \) is equal to exactly 0 plus the probability that \( x \) is equal to exactly 1 plus the probability that \( x \) is equal to exactly 2.

You're just stacking these things on top of each other. Why are you adding them? Because this is basically an or probability. Right? This is the probability that either 0 customers open up an email or 1 or 2.

And we add when we do or probabilities, we add them together. That's all that's going on here. Alright? So I'm going to go ahead and set up these equations over here, and then I'm just going to give you the numbers. So \( px = 0 \), again, using this formula over here and using our \( np \) and \( q \), it's just going to become \( 10 \choose 0 \), then this is going to be 0.4 to the \( 0 \) power, then sorry, \( 0.4 \times 0.6 \) raised to the \( 10 \) power. Alright? So remember, the exponent should always equal the number of trials that you have. So now we have plus. This is going to be \( 10 \choose 1 \).

Now this is going to be 0.4 raised to the 1 and then times 0.6 raised to the 9 power. Alright? Plus, and now \( 10 \choose 2 \), then we have 0.4 raised to the 2 power, and then times \( 0.6 \) raised to the 8. Alright? So go ahead and pause the video and see if you guys can work these out on your own.

But if not, what we could do, I'm just going to give you the numbers for these three values. Okay. Okay. So I'm going to keep it to 3 decimal places. This is going to be 0.006.

The second one is going to be 0.040. That's the second number. The third one is going to be 0.121. Alright? That makes sense because basically, what happens is if 40% is sort of the most probable, the lower numbers are going to have very low probabilities, and they'll kind of increase up until a certain point.

Alright? So this makes sense. And basically, if you just add up all of these three numbers, what you should get is a total probability of 0.167. That's our first answer. So in other words, there's a 16.7% chance that between 0 and 2 customers will open up this email.

Alright? That's really all there is to it. You're just adding up a bunch of these things. Let's take a look at the second example now. This is example B.

So what's now the probability that at least 3 open up the email? Alright? So again, you can sort of condense this down into sort of a shorthand. At least 3, if you go up to our table, basically just means that you're going to start at that lower number. So that number that they're giving you, 3, and then you're going to go up and add up everything until you have the maximum number of successes, which is, in this case, 10.

So at least 3. So in other words, at least 3, which is \( p \) that \( x \) is greater than or equal to 3, is basically that \( p(x = 3) \) all the way up until, and you're going to add up a bunch of things all the way up until \( p(x = 10) \). Alright? So in other words, if we were to expand this out, we're going to have something like 7 terms in this. So unfortunately, what happens is this is very long.

We'd have to do exactly what we did over here, but we'd have to do it for 7 or 8 terms instead of only 3. Is there a faster way to do this? And actually, there is, because we could use what's called the complement. Remember, the complement is basically just kind of like the opposite. Right?

So here's one way to think about this. Right? So \( p(x \geq 3) \) is kind of the opposite of \( p(x \leq 2) \). Alright? We know that the sum total of all of the probabilities of all of the outcomes in this experiment has to total up to 1.

So \( p(x \geq 3) \) customers opening up the email, plus \( p(x \leq 2) \), all of those probabilities have to add up to 1. We already just found what \( p \leq 2 \) is because that's basically what these three terms represent. Between 0 and 2 is another way of saying \( p \) is less than or equal to 2, right, because you're going all the way down to 0. So basically, what you can do here is that \( p(x \geq 3) \) is equal to 1 minus \( p_0 + p_1 + p_2 \). And we actually already know what all of these answers, what all of these numbers are.

Alright? So what I can do here is even if I hadn't calculated this in part A, this would still be a much faster way to do this problem, and I wouldn't actually have to crank out, like, 8 different terms because I only have 3 terms to figure out instead of 8. So sometimes it may be easier to subtract the complements, the probability of the complement, rather than adding up a bunch of probabilities. Alright? That's the point here.

Okay. So we actually already know what all of these three numbers will total up to, which is just 0.167. So basically, the probability that at least 3 customers open up the email is just going to be \( 1 - 0.167 \). So if you work this out, you're going to get 0.833. So there's an 83.3 percent chance that at least 3 customers open up this email, and that's the answer.

Alright, folks. As I do these types of problems, let me know if you have any questions, and I'll see you in the next one.

A gardener plants 8 seeds, and each has a 65% probability of germinating successfully. Find the probability that less than 4 seeds germinate successfully.

• A. 0.0266
• B. 0.0252
• C. 0.1062
• D. 0.0084