Sentence
stringlengths 102
4.09k
| video_title
stringlengths 27
104
|
|---|---|
Now, what about the normal condition? The normal condition tells us that the expected number of successes, which would be our sample size times the true proportion, and the number of failures, sample size times one minus p, need to be at least equal to 10. So they need to be greater than or equal to 10. Now, what are they for this particular scenario? Well, n is equal to 10, n is equal to 10, and our true proportion, remember, we're going to assume, when we do the significance test, we assume the null hypothesis is true, and the null hypothesis tells us that our true proportion is 0.1. So this is 0.1. This is one minus 0.1, which is 0.9. | Conditions for a z test about a proportion AP Statistics Khan Academy.mp3 |
Now, what are they for this particular scenario? Well, n is equal to 10, n is equal to 10, and our true proportion, remember, we're going to assume, when we do the significance test, we assume the null hypothesis is true, and the null hypothesis tells us that our true proportion is 0.1. So this is 0.1. This is one minus 0.1, which is 0.9. Well, 10 times 0.1 is one, so that's not greater than or equal to 10, so just off of that, we don't meet the normal condition, but even the second one, 10 times 0.9 is nine. That's also not greater than or equal to 10, so we don't meet this normal condition. We can't feel good that the sampling distribution is roughly normal, which we normally assume when we're trying to make this type of calculation. | Conditions for a z test about a proportion AP Statistics Khan Academy.mp3 |
This is one minus 0.1, which is 0.9. Well, 10 times 0.1 is one, so that's not greater than or equal to 10, so just off of that, we don't meet the normal condition, but even the second one, 10 times 0.9 is nine. That's also not greater than or equal to 10, so we don't meet this normal condition. We can't feel good that the sampling distribution is roughly normal, which we normally assume when we're trying to make this type of calculation. And then last but not least, independence. Independence is to feel good that each of the data points in your sample are independent. The results of whether they are a success or a failure is independent of each other. | Conditions for a z test about a proportion AP Statistics Khan Academy.mp3 |
We can't feel good that the sampling distribution is roughly normal, which we normally assume when we're trying to make this type of calculation. And then last but not least, independence. Independence is to feel good that each of the data points in your sample are independent. The results of whether they are a success or a failure is independent of each other. Now, if she was surveying these people with replacement, if each data point was with replacement, you would definitely meet this independence condition, but she didn't do it with replacement, but there's another way to go about it. You could use your 10% rule. If your sample size is less than 10% of the population size, then it's okay. | Conditions for a z test about a proportion AP Statistics Khan Academy.mp3 |
Sunil and his friends have been using a group messaging app for over a year to chat with each other. He suspects that on average, they send each other more than 100 messages per day. Sunil takes a random sample of seven days from their chat history and records how many messages were sent on those days. The sample data are strongly skewed to the right with a mean of 125 messages and a standard deviation of 44 messages. He wants to use these sample data to conduct a t-test about the mean. Which conditions for performing this type of significance test have been met? So let's just think about what's going on here. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
The sample data are strongly skewed to the right with a mean of 125 messages and a standard deviation of 44 messages. He wants to use these sample data to conduct a t-test about the mean. Which conditions for performing this type of significance test have been met? So let's just think about what's going on here. Sunil might have some type of a null hypothesis. Maybe he got this 100, maybe he read a magazine article that says that on average, the average teenager sends 100 text messages per day. And so maybe the null hypothesis is that the mean amount of messages per day that he and his friends send, which was signified by mu, maybe the null is 100, that they're no different than all other teenagers, and maybe he suspects, and actually they say it right over here, his alternative hypothesis would be what he suspects, that they'd send more than 100 text messages per day. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
So let's just think about what's going on here. Sunil might have some type of a null hypothesis. Maybe he got this 100, maybe he read a magazine article that says that on average, the average teenager sends 100 text messages per day. And so maybe the null hypothesis is that the mean amount of messages per day that he and his friends send, which was signified by mu, maybe the null is 100, that they're no different than all other teenagers, and maybe he suspects, and actually they say it right over here, his alternative hypothesis would be what he suspects, that they'd send more than 100 text messages per day. And so what he does is he takes a sample from the population of days, and there's over 365. They say they've been using the group messaging app for over a year. And he takes seven of those days. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
And so maybe the null hypothesis is that the mean amount of messages per day that he and his friends send, which was signified by mu, maybe the null is 100, that they're no different than all other teenagers, and maybe he suspects, and actually they say it right over here, his alternative hypothesis would be what he suspects, that they'd send more than 100 text messages per day. And so what he does is he takes a sample from the population of days, and there's over 365. They say they've been using the group messaging app for over a year. And he takes seven of those days. So n is equal to seven. And from that, he calculates sample statistics. He calculates the sample mean, which is trying to estimate the true population mean right over here. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
And he takes seven of those days. So n is equal to seven. And from that, he calculates sample statistics. He calculates the sample mean, which is trying to estimate the true population mean right over here. And he also is able to calculate a sample standard deviation. And what you do in a significance test is you say, well, what is the probability of getting this sample mean or something even more extreme, assuming the null hypothesis? And if that probability is below a preset threshold, then you would reject the null hypothesis, and it would suggest the alternative. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
He calculates the sample mean, which is trying to estimate the true population mean right over here. And he also is able to calculate a sample standard deviation. And what you do in a significance test is you say, well, what is the probability of getting this sample mean or something even more extreme, assuming the null hypothesis? And if that probability is below a preset threshold, then you would reject the null hypothesis, and it would suggest the alternative. But in order to feel good about that significance test and be able to even calculate that p-value with confidence, there are conditions for performing this type of significance test. The first is is that this is truly a random sample, and that's known as the random condition. And you have seen this before when we did significance tests with proportions. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
And if that probability is below a preset threshold, then you would reject the null hypothesis, and it would suggest the alternative. But in order to feel good about that significance test and be able to even calculate that p-value with confidence, there are conditions for performing this type of significance test. The first is is that this is truly a random sample, and that's known as the random condition. And you have seen this before when we did significance tests with proportions. Here we're doing it with means, population mean, sample mean. In the past, we did it with population proportion and sample proportion. Well, the random condition, it says it right here, Sunil takes a random sample of seven days from their chat history. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
And you have seen this before when we did significance tests with proportions. Here we're doing it with means, population mean, sample mean. In the past, we did it with population proportion and sample proportion. Well, the random condition, it says it right here, Sunil takes a random sample of seven days from their chat history. They don't say how he did it, but we'll just take their word for it that it was a random sample. The next condition is sometimes known as the independence, independence condition. And that's that the individual observations in our sample are roughly independent. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
Well, the random condition, it says it right here, Sunil takes a random sample of seven days from their chat history. They don't say how he did it, but we'll just take their word for it that it was a random sample. The next condition is sometimes known as the independence, independence condition. And that's that the individual observations in our sample are roughly independent. One way that they would be independent for sure is if Sunil is sampling with replacement. They don't say that, but another condition, so you either could have replacement, sampling with replacement, or another way where you could feel that it's roughly independent is if your sample size is less than or equal to 10% of the population. Now in this situation, he took seven, he took a sample size of seven, and then the population of days, it says that they've been using the group messaging app for over a year, so they've been using it for over 365 days. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
And that's that the individual observations in our sample are roughly independent. One way that they would be independent for sure is if Sunil is sampling with replacement. They don't say that, but another condition, so you either could have replacement, sampling with replacement, or another way where you could feel that it's roughly independent is if your sample size is less than or equal to 10% of the population. Now in this situation, he took seven, he took a sample size of seven, and then the population of days, it says that they've been using the group messaging app for over a year, so they've been using it for over 365 days. So seven is for sure less than or equal to 10% of 365, which would be 36.5. So we meet this condition, which allows us to meet the independence condition. Now the last condition is often known as the normal condition, and this is to feel good that the sampling distribution of the sample means right over here is approximately normal. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
Now in this situation, he took seven, he took a sample size of seven, and then the population of days, it says that they've been using the group messaging app for over a year, so they've been using it for over 365 days. So seven is for sure less than or equal to 10% of 365, which would be 36.5. So we meet this condition, which allows us to meet the independence condition. Now the last condition is often known as the normal condition, and this is to feel good that the sampling distribution of the sample means right over here is approximately normal. And this is going to be a little bit different than what we saw with significance tests when we dealt with proportions. There's a few ways to feel good that the sampling distribution of the sample means is normal. One is is if the underlying parent population normal. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
Now the last condition is often known as the normal condition, and this is to feel good that the sampling distribution of the sample means right over here is approximately normal. And this is going to be a little bit different than what we saw with significance tests when we dealt with proportions. There's a few ways to feel good that the sampling distribution of the sample means is normal. One is is if the underlying parent population normal. So parent, parent population normal. Now they don't tell us anything that there's actually a normal distribution for the amount of time that they spend on a given day. So we don't know this one for sure, but sometimes you might. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
One is is if the underlying parent population normal. So parent, parent population normal. Now they don't tell us anything that there's actually a normal distribution for the amount of time that they spend on a given day. So we don't know this one for sure, but sometimes you might. Another way is to feel good that our sample size is greater than or equal to 30. And this comes from the central limit theorem that then our sampling distribution is going to be roughly normal. But we see very clearly our sample size is not greater than or equal to 30, so we don't meet that constraint either. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
So we don't know this one for sure, but sometimes you might. Another way is to feel good that our sample size is greater than or equal to 30. And this comes from the central limit theorem that then our sampling distribution is going to be roughly normal. But we see very clearly our sample size is not greater than or equal to 30, so we don't meet that constraint either. Now the third way that we could feel good that our sampling distribution of our sample mean is roughly normal is if our sample, is if our sample is symmetric, symmetric, and there are no outliers, or maybe even you could say no significant outliers. Now is this the case? Well it says right over here, the sample data are strongly skewed to the right with a mean of 125 messages and a standard deviation of 44 messages. | Conditions for a t test about a mean AP Statistics Khan Academy.mp3 |
Suppose that Erica simultaneously rolls a six-sided die and a four-sided die. Let A be the event that she rolls doubles. Let me write this. A be the event that she rolls doubles and B be the event that the four-sided die is a four. Use the sample space of possible outcomes below to answer each of the following questions. Fair enough. What is probability of A, the probability that Erica rolls doubles? | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
A be the event that she rolls doubles and B be the event that the four-sided die is a four. Use the sample space of possible outcomes below to answer each of the following questions. Fair enough. What is probability of A, the probability that Erica rolls doubles? Well, over here we have our sample space of possible outcomes. Each of these are equally likely. And so let's see how many of them there are. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
What is probability of A, the probability that Erica rolls doubles? Well, over here we have our sample space of possible outcomes. Each of these are equally likely. And so let's see how many of them there are. There are one, two, three, four by one, two, three, four, five, six. So there are 24 possible outcomes, which makes sense. There's four possible outcomes for the four-sided die and six possible outcomes for the six-sided die. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
And so let's see how many of them there are. There are one, two, three, four by one, two, three, four, five, six. So there are 24 possible outcomes, which makes sense. There's four possible outcomes for the four-sided die and six possible outcomes for the six-sided die. So you have a total of 24 equally likely outcomes. So probability of, let me write it here. So probability of A, probability of A is going to be the fraction of the 24 equally likely outcomes that involve event A, that she rolls doubles. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
There's four possible outcomes for the four-sided die and six possible outcomes for the six-sided die. So you have a total of 24 equally likely outcomes. So probability of, let me write it here. So probability of A, probability of A is going to be the fraction of the 24 equally likely outcomes that involve event A, that she rolls doubles. So let's think about that. This is, she has rolled doubles, one and a one. They don't look the same, but they're both ones. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So probability of A, probability of A is going to be the fraction of the 24 equally likely outcomes that involve event A, that she rolls doubles. So let's think about that. This is, she has rolled doubles, one and a one. They don't look the same, but they're both ones. Let's see, we have a two and a two. We have a three and a three. And we have a four and a four. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
They don't look the same, but they're both ones. Let's see, we have a two and a two. We have a three and a three. And we have a four and a four. And it's impossible to have a five and a five because the four-sided die only goes up to four. So there's four possibilities, four of the 24 equally likely possibilities involve rolling doubles. So there's a 424th probability, or if we divide the numerator and denominator by four, it is a 1 6th probability that Erica, a 1 6th probability that Erica rolls doubles. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
And we have a four and a four. And it's impossible to have a five and a five because the four-sided die only goes up to four. So there's four possibilities, four of the 24 equally likely possibilities involve rolling doubles. So there's a 424th probability, or if we divide the numerator and denominator by four, it is a 1 6th probability that Erica, a 1 6th probability that Erica rolls doubles. What is probability of B? The probability that the four-sided die is a four. So the probability of B, well, once again, there's 24 equally likely possibilities. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So there's a 424th probability, or if we divide the numerator and denominator by four, it is a 1 6th probability that Erica, a 1 6th probability that Erica rolls doubles. What is probability of B? The probability that the four-sided die is a four. So the probability of B, well, once again, there's 24 equally likely possibilities. And how many of them involve the four-sided die being a four? Well, you have all of these right over here involve a four-sided die being a four. So this is one, two, three, four, five, six of the 24 equally likely possibilities. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So the probability of B, well, once again, there's 24 equally likely possibilities. And how many of them involve the four-sided die being a four? Well, you have all of these right over here involve a four-sided die being a four. So this is one, two, three, four, five, six of the 24 equally likely possibilities. Or you could say 1 4th of the equally likely possibilities, or the probability is 1 4th, which makes sense because probability of B, it kind of ignores the six-sided die. And it just says, well, what's the probability that the four-sided die is four? Well, that's one out of the four possible outcomes for that four-sided die. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So this is one, two, three, four, five, six of the 24 equally likely possibilities. Or you could say 1 4th of the equally likely possibilities, or the probability is 1 4th, which makes sense because probability of B, it kind of ignores the six-sided die. And it just says, well, what's the probability that the four-sided die is four? Well, that's one out of the four possible outcomes for that four-sided die. What is the probability of A given B? The probability that Erica rolls doubles given that the four-sided die is a four. So let's just think about this a little bit. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Well, that's one out of the four possible outcomes for that four-sided die. What is the probability of A given B? The probability that Erica rolls doubles given that the four-sided die is a four. So let's just think about this a little bit. Probability of A given that B has happened. Given that B has happened. So essentially, we are restricting our equally likely possibilities now to the situation where B has happened. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So let's just think about this a little bit. Probability of A given that B has happened. Given that B has happened. So essentially, we are restricting our equally likely possibilities now to the situation where B has happened. Given B means we're assuming that B has happened. So now we're restricting our sample space of possible outcomes where B has happened to this right over here. So now there are one, two, three, four, five, six equally likely outcomes. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So essentially, we are restricting our equally likely possibilities now to the situation where B has happened. Given B means we're assuming that B has happened. So now we're restricting our sample space of possible outcomes where B has happened to this right over here. So now there are one, two, three, four, five, six equally likely outcomes. And how many of them involve A happening? Well, this one right over here that we had already circled. This is the one out of the six equally likely outcomes that involve doubles. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So now there are one, two, three, four, five, six equally likely outcomes. And how many of them involve A happening? Well, this one right over here that we had already circled. This is the one out of the six equally likely outcomes that involve doubles. So there's a 1 6th probability. Now that makes sense. Let me just write this down. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
This is the one out of the six equally likely outcomes that involve doubles. So there's a 1 6th probability. Now that makes sense. Let me just write this down. This is one over six. Why does this make sense? Because with a four-sided die, we're assuming is a four. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Let me just write this down. This is one over six. Why does this make sense? Because with a four-sided die, we're assuming is a four. So it's essentially, this is analogous to saying when you roll a six-sided die, what's the probability that you get a four as well? Because that's the only way you're going to get doubles given that the four-sided die is four. And we see that right over here. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Because with a four-sided die, we're assuming is a four. So it's essentially, this is analogous to saying when you roll a six-sided die, what's the probability that you get a four as well? Because that's the only way you're going to get doubles given that the four-sided die is four. And we see that right over here. The six-sided die has to be a four as well in order for this to be doubles because we're assuming where it's given that B, we're given of event B. We're restricting our sample space with event B. What is the probability of B given A? | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
And we see that right over here. The six-sided die has to be a four as well in order for this to be doubles because we're assuming where it's given that B, we're given of event B. We're restricting our sample space with event B. What is the probability of B given A? The probability that the four-sided die is four given that Erica rolls doubles. So let's just think about that a little bit. So the probability of B given A. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
What is the probability of B given A? The probability that the four-sided die is four given that Erica rolls doubles. So let's just think about that a little bit. So the probability of B given A. B given that A is true. So what's this going to be? Well, we've already, so this means we're going to restrict our sample space to the essentially four equally likely outcomes that A has happened. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So the probability of B given A. B given that A is true. So what's this going to be? Well, we've already, so this means we're going to restrict our sample space to the essentially four equally likely outcomes that A has happened. So where A is true, I guess I could say. So there's one, two, three, four. And how many of them involve event B being true? | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Well, we've already, so this means we're going to restrict our sample space to the essentially four equally likely outcomes that A has happened. So where A is true, I guess I could say. So there's one, two, three, four. And how many of them involve event B being true? Well, the only one of these four that involve event B being true is this one right over here where we've got our doubles. So there is a 1 1 4th probability that if we assume, given that we've gotten doubles, the probability that the four-sided die is a four. So this is a 1 1 4th probability. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
And how many of them involve event B being true? Well, the only one of these four that involve event B being true is this one right over here where we've got our doubles. So there is a 1 1 4th probability that if we assume, given that we've gotten doubles, the probability that the four-sided die is a four. So this is a 1 1 4th probability. And that makes sense. If we've got doubles, and one of them's a four-sided die, we either have doubles at one, doubles at two, doubles at three, or doubles at four. You see that here, doubles one, doubles two, doubles three, doubles four. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So this is a 1 1 4th probability. And that makes sense. If we've got doubles, and one of them's a four-sided die, we either have doubles at one, doubles at two, doubles at three, or doubles at four. You see that here, doubles one, doubles two, doubles three, doubles four. Well, given that, what's the probability that the four-sided die is a four? Well, that means that's one of these four outcomes where it's a double fours right over here. All right, what is the probability of A and B? | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
You see that here, doubles one, doubles two, doubles three, doubles four. Well, given that, what's the probability that the four-sided die is a four? Well, that means that's one of these four outcomes where it's a double fours right over here. All right, what is the probability of A and B? The probability that Erica rolls doubles and the second die is four. So this means both A and B happened. Well, let's look at this. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
All right, what is the probability of A and B? The probability that Erica rolls doubles and the second die is four. So this means both A and B happened. Well, let's look at this. Actually, let me write it here. Let me do it in this, let me do it in a new color. So the probability of, I'll write and here in a neutral color, probability of A and B, probability of A and B is equal to, well, now we're looking at, once again, we have 24 equally likely outcomes. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Well, let's look at this. Actually, let me write it here. Let me do it in this, let me do it in a new color. So the probability of, I'll write and here in a neutral color, probability of A and B, probability of A and B is equal to, well, now we're looking at, once again, we have 24 equally likely outcomes. We have 24 equally likely outcomes. And how many of them involve A and B? Well, to get A and B, you have to have doubles and the four-sided die needs to be a four. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So the probability of, I'll write and here in a neutral color, probability of A and B, probability of A and B is equal to, well, now we're looking at, once again, we have 24 equally likely outcomes. We have 24 equally likely outcomes. And how many of them involve A and B? Well, to get A and B, you have to have doubles and the four-sided die needs to be a four. Essentially, you have to have doubles four. Well, there's only one outcome out of the 24 equally likely outcomes that meets that situation, this one right over here. So there is a 1 4th, sorry, 1 24th probability. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Well, to get A and B, you have to have doubles and the four-sided die needs to be a four. Essentially, you have to have doubles four. Well, there's only one outcome out of the 24 equally likely outcomes that meets that situation, this one right over here. So there is a 1 4th, sorry, 1 24th probability. So one over 24. What is probability of A times probability of B given A? Well, here we could just go back to our numbers right over here. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So there is a 1 4th, sorry, 1 24th probability. So one over 24. What is probability of A times probability of B given A? Well, here we could just go back to our numbers right over here. Probability of A, that's going to be one over six. Let me do that in a magenta color. I like to keep my colors, be careful about my colors. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Well, here we could just go back to our numbers right over here. Probability of A, that's going to be one over six. Let me do that in a magenta color. I like to keep my colors, be careful about my colors. That's 1 6th times probability of B given A. So probability of B given A is 1 4th right over here, times 1 4th, which is, curious enough, 1 24th. 1 24th. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
I like to keep my colors, be careful about my colors. That's 1 6th times probability of B given A. So probability of B given A is 1 4th right over here, times 1 4th, which is, curious enough, 1 24th. 1 24th. What is probability of B times probability of A given B? Well, probability of B, we figured out, is 1 4th. 1 4th. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
1 24th. What is probability of B times probability of A given B? Well, probability of B, we figured out, is 1 4th. 1 4th. And the probability of A given B is 1 6th, times 1 6th, times 1 6th, which is equal to 1 24th. Now, does it make sense that the probability of A and B is 1 24th, that the probability of A times probability of B given A is 1 24th, and the probability of B times probability of A given B, they're all 1 24th. Is this always going to be the case? | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
1 4th. And the probability of A given B is 1 6th, times 1 6th, times 1 6th, which is equal to 1 24th. Now, does it make sense that the probability of A and B is 1 24th, that the probability of A times probability of B given A is 1 24th, and the probability of B times probability of A given B, they're all 1 24th. Is this always going to be the case? Well, sure. Think about what probability of A and B means. Well, it means that they both happened. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Is this always going to be the case? Well, sure. Think about what probability of A and B means. Well, it means that they both happened. But that's the same way as saying, well, what's the probability of, let's just say A is happening. Well, now for B and A to happen, it's just going to be that times the probability that B is true given that A is true, because you could kind of say, well, we're already kind of constraining it, we're already multiplying by the probability of A being true, and now we're multiplying by the probability that B is true given A is true. I actually often like to swap these around, just it gets a little bit clearer in my head. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
Well, it means that they both happened. But that's the same way as saying, well, what's the probability of, let's just say A is happening. Well, now for B and A to happen, it's just going to be that times the probability that B is true given that A is true, because you could kind of say, well, we're already kind of constraining it, we're already multiplying by the probability of A being true, and now we're multiplying by the probability that B is true given A is true. I actually often like to swap these around, just it gets a little bit clearer in my head. So this one, let's just write it like this. The probability of B given A times the probability of A. So this is the probability that event A is true, and this is the probability that B, event B is true, given that we know that A is true. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
I actually often like to swap these around, just it gets a little bit clearer in my head. So this one, let's just write it like this. The probability of B given A times the probability of A. So this is the probability that event A is true, and this is the probability that B, event B is true, given that we know that A is true. And it completely makes sense that this is going to be the same thing as the probability of A and B. Clearly, this is the probability of both of these, both A and B happening, and you can go the other way around. The probability of A given B times the probability of B, that would also be, so we're saying B needs to be true, and that given that B is true, that A needs to be true as well. | Analyzing dependent probability Probability and Statistics Khan Academy.mp3 |
So for each of those scenarios, four different people could sit in chair two. Now, for each of these scenarios now, so we have 20 scenarios, five times four, we have 20 scenarios where we've seated seat one and seat two. How many people could we now seat in seat three for each of those 20 scenarios? Well, three people haven't sat down yet, so there's three possibilities there. So now there's five times four times three scenarios for seating the first three people. How many people are left for seat four? Well, two people haven't sat down yet, so there's two possibilities. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Well, three people haven't sat down yet, so there's three possibilities there. So now there's five times four times three scenarios for seating the first three people. How many people are left for seat four? Well, two people haven't sat down yet, so there's two possibilities. So now there's five times four times three times two scenarios of seating the first four seats. And for each of those, how many possibilities are there for the fifth seat? Well, for each of those scenarios, we only have one person who hasn't sat down left, so there's one possibility. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Well, two people haven't sat down yet, so there's two possibilities. So now there's five times four times three times two scenarios of seating the first four seats. And for each of those, how many possibilities are there for the fifth seat? Well, for each of those scenarios, we only have one person who hasn't sat down left, so there's one possibility. And so the number of permutations, the number of, let me write this down, the number of permutations, permutations of seating these five people in five chairs is five factorial. Five factorial, which is equal to five times four times three times two times one, which of course is equal to, let's see, 20 times six, which is equal to 120. And we've already covered this in a previous video. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Well, for each of those scenarios, we only have one person who hasn't sat down left, so there's one possibility. And so the number of permutations, the number of, let me write this down, the number of permutations, permutations of seating these five people in five chairs is five factorial. Five factorial, which is equal to five times four times three times two times one, which of course is equal to, let's see, 20 times six, which is equal to 120. And we've already covered this in a previous video. But now let's do something maybe more interesting, or maybe you might find it less interesting. Let's say that we still have, let's still say we have these five people, but we don't have as many chairs, so not everyone is going to be able to sit down. So let's say that we only have three chairs. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
And we've already covered this in a previous video. But now let's do something maybe more interesting, or maybe you might find it less interesting. Let's say that we still have, let's still say we have these five people, but we don't have as many chairs, so not everyone is going to be able to sit down. So let's say that we only have three chairs. So we have chair one, we have chair two, and we have chair three. So how many ways can you have five people where only three of them are going to sit down in these three chairs? And we care which chair they sit in, and I encourage you to pause the video and think about it. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So let's say that we only have three chairs. So we have chair one, we have chair two, and we have chair three. So how many ways can you have five people where only three of them are going to sit down in these three chairs? And we care which chair they sit in, and I encourage you to pause the video and think about it. So I am assuming you have had your go at it. So let's use the same logic. So how many, if we seat them in order, we might as well, how many different people, if we haven't sat anyone yet, how many different people could sit in seat one? | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
And we care which chair they sit in, and I encourage you to pause the video and think about it. So I am assuming you have had your go at it. So let's use the same logic. So how many, if we seat them in order, we might as well, how many different people, if we haven't sat anyone yet, how many different people could sit in seat one? Well, we could have, if no one sat down, we have five different people. Well, five different people could sit in seat one. Well, for each of these scenarios where one person has already sat in seat one, how many people could sit in seat two? | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So how many, if we seat them in order, we might as well, how many different people, if we haven't sat anyone yet, how many different people could sit in seat one? Well, we could have, if no one sat down, we have five different people. Well, five different people could sit in seat one. Well, for each of these scenarios where one person has already sat in seat one, how many people could sit in seat two? Well, in each of these scenarios, if one person has sat down, there's four people left who haven't been seated, so four people could sit in seat two. So we have five times four scenarios where we have seated seats one and seat two. Now, for each of those 20 scenarios, how many people could sit in seat three? | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Well, for each of these scenarios where one person has already sat in seat one, how many people could sit in seat two? Well, in each of these scenarios, if one person has sat down, there's four people left who haven't been seated, so four people could sit in seat two. So we have five times four scenarios where we have seated seats one and seat two. Now, for each of those 20 scenarios, how many people could sit in seat three? Well, we haven't sat, we haven't, we haven't seaten or sat three of the people yet, so for each of these 20, we could put three different people in seat three. So that gives us five times four times three scenarios. So this is equal to five times four times three scenarios, which is equal to, this is equal to 60. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Now, for each of those 20 scenarios, how many people could sit in seat three? Well, we haven't sat, we haven't, we haven't seaten or sat three of the people yet, so for each of these 20, we could put three different people in seat three. So that gives us five times four times three scenarios. So this is equal to five times four times three scenarios, which is equal to, this is equal to 60. So there's 60 permutations of sitting five people in three chairs. Now, this, and this is my brain, you know, whenever I start to think in terms of permutations, I actually think in these ways. I just literally draw it out because especially, you know, I don't like formulas. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So this is equal to five times four times three scenarios, which is equal to, this is equal to 60. So there's 60 permutations of sitting five people in three chairs. Now, this, and this is my brain, you know, whenever I start to think in terms of permutations, I actually think in these ways. I just literally draw it out because especially, you know, I don't like formulas. I like to actually conceptualize and visualize what I'm doing. But you might say, hey, you know, when we just did five different people in five different chairs and we cared which seat they sit in, we had this five factorial. Well, you know, factorial is kind of a neat little operation. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
I just literally draw it out because especially, you know, I don't like formulas. I like to actually conceptualize and visualize what I'm doing. But you might say, hey, you know, when we just did five different people in five different chairs and we cared which seat they sit in, we had this five factorial. Well, you know, factorial is kind of a neat little operation. How can I relate factorial to what we did just now? Well, it looks like we kind of did factorial, but then we stopped. We stopped at, we didn't go times two times one. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Well, you know, factorial is kind of a neat little operation. How can I relate factorial to what we did just now? Well, it looks like we kind of did factorial, but then we stopped. We stopped at, we didn't go times two times one. So one way to think about what we just did is we just did five times four times three times two times one. But of course, we actually didn't do the two times one. So you could take that and you could divide by two times one. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
We stopped at, we didn't go times two times one. So one way to think about what we just did is we just did five times four times three times two times one. But of course, we actually didn't do the two times one. So you could take that and you could divide by two times one. And if you did that, then this two times one would cancel with that two times one and you'd be left with five times four times three. And the whole reason I'm writing this way is that now I could write it in terms of factorial. I could write this as five factorial over two factorial. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So you could take that and you could divide by two times one. And if you did that, then this two times one would cancel with that two times one and you'd be left with five times four times three. And the whole reason I'm writing this way is that now I could write it in terms of factorial. I could write this as five factorial over two factorial. But then you might have the question, well, where did this two come from? You know, I have three seats. Where did this two come from? | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
I could write this as five factorial over two factorial. But then you might have the question, well, where did this two come from? You know, I have three seats. Where did this two come from? Well, think about it. I multiplied five times four times three. I kept going until I had that many seats and then I didn't do the remainder. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Where did this two come from? Well, think about it. I multiplied five times four times three. I kept going until I had that many seats and then I didn't do the remainder. So the number of, so the things that I left out, the things that I left out, that was essentially the number of people minus the number of chairs. So I was trying to put five things in three places. So five minus three, that gave me two left over. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
I kept going until I had that many seats and then I didn't do the remainder. So the number of, so the things that I left out, the things that I left out, that was essentially the number of people minus the number of chairs. So I was trying to put five things in three places. So five minus three, that gave me two left over. So I could write it like this. I could write it as five, let me use those same colors. I could write it as five factorial over, over five minus three, which of course is two. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So five minus three, that gave me two left over. So I could write it like this. I could write it as five, let me use those same colors. I could write it as five factorial over, over five minus three, which of course is two. Five minus three factorial. And so another way of thinking about it, if we wanted to generalize, is if you're trying to put, if you're trying to figure out the number of permutations, and there's a bunch of notations for writing this. If you're trying to figure out the number of permutations where you could put n people in r seats, or the number of permutations, you could put n people in r seats, and there's other notations as well. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
I could write it as five factorial over, over five minus three, which of course is two. Five minus three factorial. And so another way of thinking about it, if we wanted to generalize, is if you're trying to put, if you're trying to figure out the number of permutations, and there's a bunch of notations for writing this. If you're trying to figure out the number of permutations where you could put n people in r seats, or the number of permutations, you could put n people in r seats, and there's other notations as well. Well this is just going to be n factorial over n minus r factorial. Here n was five, r was three, and five minus three is two. Now, you'll see this in a probability or statistics class, and people might memorize this thing, it seems like this kind of daunting thing. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
If you're trying to figure out the number of permutations where you could put n people in r seats, or the number of permutations, you could put n people in r seats, and there's other notations as well. Well this is just going to be n factorial over n minus r factorial. Here n was five, r was three, and five minus three is two. Now, you'll see this in a probability or statistics class, and people might memorize this thing, it seems like this kind of daunting thing. I'll just tell you right now, the whole reason why I just showed this to you is so that you could connect it with what you might see in your textbook, or what you might see in a class, or when you see this type of formula, you see that it's not coming out of, it's not some type of voodoo magic, but I will tell you that for me, personally, I never use this formula. I always reason it through, because if you just memorize the formula, you're always going to wait, does this formula apply there? What's n, what's r? | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Now, you'll see this in a probability or statistics class, and people might memorize this thing, it seems like this kind of daunting thing. I'll just tell you right now, the whole reason why I just showed this to you is so that you could connect it with what you might see in your textbook, or what you might see in a class, or when you see this type of formula, you see that it's not coming out of, it's not some type of voodoo magic, but I will tell you that for me, personally, I never use this formula. I always reason it through, because if you just memorize the formula, you're always going to wait, does this formula apply there? What's n, what's r? But if you reason it through, it comes out of straight logic. You don't have to memorize anything, you don't feel like you're just memorizing without understanding, you're just using your deductive reasoning, your logic. And that's especially valuable, because as we'll see, not every scenario is going to fit so cleanly into what we did. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
What's n, what's r? But if you reason it through, it comes out of straight logic. You don't have to memorize anything, you don't feel like you're just memorizing without understanding, you're just using your deductive reasoning, your logic. And that's especially valuable, because as we'll see, not every scenario is going to fit so cleanly into what we did. There might be some tweaks on this, where maybe only person B likes sitting in one of the chairs, or who knows what it might be, and then your formula is going to be useless. So I like reasoning through it like this, but I just showed you this so that you could connect it to a formula that you might see in a lecture, or in a class. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
And that's especially valuable, because as we'll see, not every scenario is going to fit so cleanly into what we did. There might be some tweaks on this, where maybe only person B likes sitting in one of the chairs, or who knows what it might be, and then your formula is going to be useless. So I like reasoning through it like this, but I just showed you this so that you could connect it to a formula that you might see in a lecture, or in a class. | Permutation formula Probability and combinatorics Probability and Statistics Khan Academy.mp3 |