Sentence
stringlengths 102
4.09k
| video_title
stringlengths 27
104
|
|---|---|
But what's the probability of each of these combinations? Well, what's the probability of heads? That's 0.5 times 0.5 times 0.5 times 0.5. And then the probability of tails, since it's a fair coin, is also 0.5, times 0.5 times 0.5. So each of these, there's a 1 half chance of getting a heads times a 1 half chance of a tails times a 1 half chance of a heads times a 1 half chance of a tails, et cetera, et cetera. So each of these are essentially 1 half times 1 half, 6 times. So the probability of each of the combinations is 1 half to the 6th power. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And then the probability of tails, since it's a fair coin, is also 0.5, times 0.5 times 0.5. So each of these, there's a 1 half chance of getting a heads times a 1 half chance of a tails times a 1 half chance of a heads times a 1 half chance of a tails, et cetera, et cetera. So each of these are essentially 1 half times 1 half, 6 times. So the probability of each of the combinations is 1 half to the 6th power. And so how many combinations are there like this, where you get, out of the 6 flips, you're choosing, you're essentially choosing 4 heads. You're choosing, I'm, once again, the god of probability. I am picking 4, exactly 4, of the 6 heads. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So the probability of each of the combinations is 1 half to the 6th power. And so how many combinations are there like this, where you get, out of the 6 flips, you're choosing, you're essentially choosing 4 heads. You're choosing, I'm, once again, the god of probability. I am picking 4, exactly 4, of the 6 heads. Sorry, I'm picking 4 of exactly 6 of the flips to end up heads. I'm choosing which of the flips get selected, so to speak. So it's essentially, there are going to be, out of 6 flips, I'm choosing, as the god of probability, 4 to be heads. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
I am picking 4, exactly 4, of the 6 heads. Sorry, I'm picking 4 of exactly 6 of the flips to end up heads. I'm choosing which of the flips get selected, so to speak. So it's essentially, there are going to be, out of 6 flips, I'm choosing, as the god of probability, 4 to be heads. So that's the number of combinations, the number of unique combinations, where you have 4 out of 6 heads, times the probability of each of the combinations, which is 1 half to the 6th power. Well, let's 6 choose 4. That's 6 factorial over 4 factorial times 6 minus 4 factorial, so that's 2 factorial. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So it's essentially, there are going to be, out of 6 flips, I'm choosing, as the god of probability, 4 to be heads. So that's the number of combinations, the number of unique combinations, where you have 4 out of 6 heads, times the probability of each of the combinations, which is 1 half to the 6th power. Well, let's 6 choose 4. That's 6 factorial over 4 factorial times 6 minus 4 factorial, so that's 2 factorial. And that's times 1 half to the 6th. And I'll switch colors again, just to stop the monotony. And that equals, let's see, 6 times 5 times 4 times 3 times 2, we don't have to write the 1 times 1, I'll do it anyway. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
That's 6 factorial over 4 factorial times 6 minus 4 factorial, so that's 2 factorial. And that's times 1 half to the 6th. And I'll switch colors again, just to stop the monotony. And that equals, let's see, 6 times 5 times 4 times 3 times 2, we don't have to write the 1 times 1, I'll do it anyway. Over 4 factorial, 4 times 3 times 2 times 1, and then 2 factorial, 2 times 1. So that cancels with that. The 1 we can ignore. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And that equals, let's see, 6 times 5 times 4 times 3 times 2, we don't have to write the 1 times 1, I'll do it anyway. Over 4 factorial, 4 times 3 times 2 times 1, and then 2 factorial, 2 times 1. So that cancels with that. The 1 we can ignore. 2, divide both sides by, the numerator and denominator by 2, and this becomes a 3. So this becomes 15. So this equals 15 times 1 half to the 6th. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
The 1 we can ignore. 2, divide both sides by, the numerator and denominator by 2, and this becomes a 3. So this becomes 15. So this equals 15 times 1 half to the 6th. What's 1 half to the 6th? That's 1 over 64, right? So 1 over 64, so it becomes 15 over 64. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So this equals 15 times 1 half to the 6th. What's 1 half to the 6th? That's 1 over 64, right? So 1 over 64, so it becomes 15 over 64. So the probability of getting 4 out of 6 heads, given a fair coin, is 15 out of 64. So this is the probability of 4 out of 6 heads, given a fair coin. And if you look at it, based on our definition of B and A, this is the probability of B given A. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So 1 over 64, so it becomes 15 over 64. So the probability of getting 4 out of 6 heads, given a fair coin, is 15 out of 64. So this is the probability of 4 out of 6 heads, given a fair coin. And if you look at it, based on our definition of B and A, this is the probability of B given A. B is 4 out of 6 heads, given a fair coin. Fair enough. So let's figure out the probability of, because there's two ways of getting 4 out of 6 heads. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And if you look at it, based on our definition of B and A, this is the probability of B given A. B is 4 out of 6 heads, given a fair coin. Fair enough. So let's figure out the probability of, because there's two ways of getting 4 out of 6 heads. One, that we picked a fair coin, and then times 15 out of 64, and then there's a probability that we picked an unfair coin. So what's the probability of the unfair coin? Of getting 4 out of 6 heads, given the unfair coin? | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So let's figure out the probability of, because there's two ways of getting 4 out of 6 heads. One, that we picked a fair coin, and then times 15 out of 64, and then there's a probability that we picked an unfair coin. So what's the probability of the unfair coin? Of getting 4 out of 6 heads, given the unfair coin? Well, once again, what's the probability of each of the combinations where you get 4 out of 6? So in this situation, let's do the same one. Heads, tails, heads, tails, heads, heads. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
Of getting 4 out of 6 heads, given the unfair coin? Well, once again, what's the probability of each of the combinations where you get 4 out of 6? So in this situation, let's do the same one. Heads, tails, heads, tails, heads, heads. That's 4 out of 6 heads. But in this situation, it's not a 50% chance of getting heads, it's 80%. So it would be 0.8 times 0.2 times 0.8 times 0.2 times 0.8 times 0.8. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
Heads, tails, heads, tails, heads, heads. That's 4 out of 6 heads. But in this situation, it's not a 50% chance of getting heads, it's 80%. So it would be 0.8 times 0.2 times 0.8 times 0.2 times 0.8 times 0.8. Essentially, we have this multiplication. We can rearrange it, because it doesn't matter what order you multiply things in. So it's 0.8 to the fourth power times 0.2 squared. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So it would be 0.8 times 0.2 times 0.8 times 0.2 times 0.8 times 0.8. Essentially, we have this multiplication. We can rearrange it, because it doesn't matter what order you multiply things in. So it's 0.8 to the fourth power times 0.2 squared. And it doesn't matter. Any of the unique combinations will each have the same probability, because we can just rearrange the order in which we multiply. And then how many of these combinations are there? | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So it's 0.8 to the fourth power times 0.2 squared. And it doesn't matter. Any of the unique combinations will each have the same probability, because we can just rearrange the order in which we multiply. And then how many of these combinations are there? If we are, once again, the god of probability, and out of 6 flips, we are choosing 4 that are going to end up heads. How many ways can I pick a group of 4? Well, once again, that's times 6 choose 4. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And then how many of these combinations are there? If we are, once again, the god of probability, and out of 6 flips, we are choosing 4 that are going to end up heads. How many ways can I pick a group of 4? Well, once again, that's times 6 choose 4. And we figured out what that is. 6 choose 4 is 15. So this equals 15 times 0.8 to the fourth times 0.2 squared. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
Well, once again, that's times 6 choose 4. And we figured out what that is. 6 choose 4 is 15. So this equals 15 times 0.8 to the fourth times 0.2 squared. So this is the probability of 4 out of 6 heads given an unfair coin. So what's the total probability of getting 4 out of 6 heads? Well, it's going to be the probability of getting the fair coin, which is 1 third, times the probability of getting 4 out of 6 heads given the fair coin. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So this equals 15 times 0.8 to the fourth times 0.2 squared. So this is the probability of 4 out of 6 heads given an unfair coin. So what's the total probability of getting 4 out of 6 heads? Well, it's going to be the probability of getting the fair coin, which is 1 third, times the probability of getting 4 out of 6 heads given the fair coin. And that's this, 15 over 64, plus the probability of getting an unfair coin, 2 thirds, times the probability of getting 4 out of 6 heads given the unfair coin. And that's what we figured out here. Times 15 times 0.8 to the fourth times 0.2 squared. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
Well, it's going to be the probability of getting the fair coin, which is 1 third, times the probability of getting 4 out of 6 heads given the fair coin. And that's this, 15 over 64, plus the probability of getting an unfair coin, 2 thirds, times the probability of getting 4 out of 6 heads given the unfair coin. And that's what we figured out here. Times 15 times 0.8 to the fourth times 0.2 squared. And this is the probability of getting 4 out of 6 heads. And let's figure out what that is. Well, this will cancel out with this. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
Times 15 times 0.8 to the fourth times 0.2 squared. And this is the probability of getting 4 out of 6 heads. And let's figure out what that is. Well, this will cancel out with this. This becomes 5 out of 64. That's easy enough. 2 thirds times 15, that's 10. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
Well, this will cancel out with this. This becomes 5 out of 64. That's easy enough. 2 thirds times 15, that's 10. And now we just have to figure out what that is. Let's see. I'm going to go over the time limit to see if being a YouTube partner allows me to go over the time limit. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
2 thirds times 15, that's 10. And now we just have to figure out what that is. Let's see. I'm going to go over the time limit to see if being a YouTube partner allows me to go over the time limit. Let's see. 0.8 times 0.8 is equal to. And then times 0.2 squared. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
I'm going to go over the time limit to see if being a YouTube partner allows me to go over the time limit. Let's see. 0.8 times 0.8 is equal to. And then times 0.2 squared. So times 0.2 is equal to 0.016. So that's that. And then we say times 10, right? | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And then times 0.2 squared. So times 0.2 is equal to 0.016. So that's that. And then we say times 10, right? Because 2 thirds times 15. So times 10 is equal to 16.384%. So this term right here, let me write that down. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And then we say times 10, right? Because 2 thirds times 15. So times 10 is equal to 16.384%. So this term right here, let me write that down. And I'll switch colors again. This is 0.16384. And we're going to add that to 5 divided by 64. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So this term right here, let me write that down. And I'll switch colors again. This is 0.16384. And we're going to add that to 5 divided by 64. So let's see. 5 divided by 64 is equal to 0.07, whatever, whatever. Plus 0.16384 is equal to 0.241965. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And we're going to add that to 5 divided by 64. So let's see. 5 divided by 64 is equal to 0.07, whatever, whatever. Plus 0.16384 is equal to 0.241965. So that's the probability. Not knowing which coin I picked out, that's the probability of getting 4 out of 6 heads. When you combine it, it could be 1 third chance fair, 2 thirds chance unfair. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
Plus 0.16384 is equal to 0.241965. So that's the probability. Not knowing which coin I picked out, that's the probability of getting 4 out of 6 heads. When you combine it, it could be 1 third chance fair, 2 thirds chance unfair. So that's 24.19. I'm keeping the precision just because it might come in useful later. Percent chance. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
When you combine it, it could be 1 third chance fair, 2 thirds chance unfair. So that's 24.19. I'm keeping the precision just because it might come in useful later. Percent chance. So that's the probability of B. So let's see if we can clean this up a little bit, just because I don't think we need all of this writing now. I think we're ready to substitute into our Bayes formula, which we, Bayes' theorem that we re-derived. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
Percent chance. So that's the probability of B. So let's see if we can clean this up a little bit, just because I don't think we need all of this writing now. I think we're ready to substitute into our Bayes formula, which we, Bayes' theorem that we re-derived. Recording longer videos is dangerous, because if I make a mistake, that's more time wasted. I don't want to delete anything that could be useful. OK. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
I think we're ready to substitute into our Bayes formula, which we, Bayes' theorem that we re-derived. Recording longer videos is dangerous, because if I make a mistake, that's more time wasted. I don't want to delete anything that could be useful. OK. So let's see if we can solve the probability that we picked a fair coin given that we got 4 out of 6 heads. So that is going to be equal to, by Bayes' theorem, which should make some sense to you, that is equal to the probability of B given A. So it's the probability that we get 4 out of 6 heads given a fair coin times the probability of a fair coin over the probability of getting 4 out of 6 heads either way. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
OK. So let's see if we can solve the probability that we picked a fair coin given that we got 4 out of 6 heads. So that is going to be equal to, by Bayes' theorem, which should make some sense to you, that is equal to the probability of B given A. So it's the probability that we get 4 out of 6 heads given a fair coin times the probability of a fair coin over the probability of getting 4 out of 6 heads either way. 4 out of 6 heads. So 4 out of 6 heads given a fair coin, we figured that over here, that's 15 over 64. So this equals 15 over 64. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So it's the probability that we get 4 out of 6 heads given a fair coin times the probability of a fair coin over the probability of getting 4 out of 6 heads either way. 4 out of 6 heads. So 4 out of 6 heads given a fair coin, we figured that over here, that's 15 over 64. So this equals 15 over 64. What's the probability that we picked a fair coin? Well, there's 15 coins, and 5 of them are fair, so it's 5 out of 15, so it's 1 third. And what's the probability that, in general, we picked 4 out of 6 heads? | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
So this equals 15 over 64. What's the probability that we picked a fair coin? Well, there's 15 coins, and 5 of them are fair, so it's 5 out of 15, so it's 1 third. And what's the probability that, in general, we picked 4 out of 6 heads? Well, that's this number, 0.241965. So this equals, let's see, this is equal to 5 over 64 divided by 0.241965. And what is that equal to? | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And what's the probability that, in general, we picked 4 out of 6 heads? Well, that's this number, 0.241965. So this equals, let's see, this is equal to 5 over 64 divided by 0.241965. And what is that equal to? That's 5 divided by 64 is equal to that, divided by 0.241965 is equal to 32.3%. So that's amazing, or relatively amazing. We've now, it's a little bit less than a 1 third shot that we picked the fair coin given that we got 4 out of 6 heads. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
And what is that equal to? That's 5 divided by 64 is equal to that, divided by 0.241965 is equal to 32.3%. So that's amazing, or relatively amazing. We've now, it's a little bit less than a 1 third shot that we picked the fair coin given that we got 4 out of 6 heads. And what's interesting is, the 4 out of 6 heads, it kind of decreased the probability that we got a fair coin, right? Because before having any data on what happens when we flip it, we would have had a 1 third probability, which is 33.3, right? But given that we got more heads than tails, kind of the universe of probability is telling us that, well, if you got more heads than tails, that makes it a little bit more likely that you picked the unfair coin, which is a little bit more weighted to heads. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
We've now, it's a little bit less than a 1 third shot that we picked the fair coin given that we got 4 out of 6 heads. And what's interesting is, the 4 out of 6 heads, it kind of decreased the probability that we got a fair coin, right? Because before having any data on what happens when we flip it, we would have had a 1 third probability, which is 33.3, right? But given that we got more heads than tails, kind of the universe of probability is telling us that, well, if you got more heads than tails, that makes it a little bit more likely that you picked the unfair coin, which is a little bit more weighted to heads. But it's saying it's not that much more likely, because this isn't that unusual of a result to get even with a fair coin. And so that's why it became a little bit less likely to get a fair coin. I will, actually, let me give you a bit of an intuition visually, kind of with set theory, on why that makes sense. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
But given that we got more heads than tails, kind of the universe of probability is telling us that, well, if you got more heads than tails, that makes it a little bit more likely that you picked the unfair coin, which is a little bit more weighted to heads. But it's saying it's not that much more likely, because this isn't that unusual of a result to get even with a fair coin. And so that's why it became a little bit less likely to get a fair coin. I will, actually, let me give you a bit of an intuition visually, kind of with set theory, on why that makes sense. So if we go back to Bayes' theorem, let's just say that this is our, this is the universe of all of the events, right? That's all of the universe. There's roughly a 1 third chance that I picked a fair coin. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
I will, actually, let me give you a bit of an intuition visually, kind of with set theory, on why that makes sense. So if we go back to Bayes' theorem, let's just say that this is our, this is the universe of all of the events, right? That's all of the universe. There's roughly a 1 third chance that I picked a fair coin. So roughly 1 third of this will be fair. This is fair. This is unfair. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
There's roughly a 1 third chance that I picked a fair coin. So roughly 1 third of this will be fair. This is fair. This is unfair. And then if I picked a fair coin, we figured out that it was roughly a 15 out of 64 shot that I get 4 out of 6 heads, so maybe that's this little section of the, let me do it in a different color, that's this section. And then we figured out if we have an unfair coin, I forgot what the exact number is, but there was some probability that we get 4 out of 6 heads. It's actually a little bit bigger. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
This is unfair. And then if I picked a fair coin, we figured out that it was roughly a 15 out of 64 shot that I get 4 out of 6 heads, so maybe that's this little section of the, let me do it in a different color, that's this section. And then we figured out if we have an unfair coin, I forgot what the exact number is, but there was some probability that we get 4 out of 6 heads. It's actually a little bit bigger. It's like that. So this is getting 4 out of 6 heads given you got an unfair coin. This is getting 4 out of 6 heads given that you got a fair coin. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
It's actually a little bit bigger. It's like that. So this is getting 4 out of 6 heads given you got an unfair coin. This is getting 4 out of 6 heads given that you got a fair coin. And then this whole area is a probability that you get 4 out of 6 heads. So all Bayes' theorem told us is, look, we got 4 out of 6 heads, so we're in this universe where we got 4 out of 6 heads. And if we got 4 out of 6 heads, 1 third of this universe, roughly, or 32.3% of this subset of 4 out of 6 heads, intersects with the fair coin universe. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
This is getting 4 out of 6 heads given that you got a fair coin. And then this whole area is a probability that you get 4 out of 6 heads. So all Bayes' theorem told us is, look, we got 4 out of 6 heads, so we're in this universe where we got 4 out of 6 heads. And if we got 4 out of 6 heads, 1 third of this universe, roughly, or 32.3% of this subset of 4 out of 6 heads, intersects with the fair coin universe. So this 32.3% is essentially this fraction of the total probability of getting 4 out of 6 heads. Anyway, hopefully that gave you a little bit of intuition, and I hope that YouTube lets me publish this video because I'm on my 17th minute. I'll see you in the next video. | Conditional probability and combinations Probability and Statistics Khan Academy.mp3 |
They want to test if this is convincing evidence that the mean amount for bottles in this batch is different than the target value of 500 milliliters. Let mu be the mean amount of liquid in each bottle in the batch. Write an appropriate set of hypotheses for their significance test, for the significance test that the quality-control expert is running. So pause this video and see if you can do that. All right, now let's do this together. So first, you're going to have two hypotheses. You're gonna have your null hypothesis and your alternative hypothesis. | Writing hypotheses for a significance test about a mean AP Statistics Khan Academy.mp3 |
So pause this video and see if you can do that. All right, now let's do this together. So first, you're going to have two hypotheses. You're gonna have your null hypothesis and your alternative hypothesis. Your null hypothesis is going to be a hypothesis about the population parameter that you care about, and it's going to assume kind of the status quo, no news here. And so the parameter that we care about is the mean amount of liquid in the bottles in the batch. So that's mu right over there. | Writing hypotheses for a significance test about a mean AP Statistics Khan Academy.mp3 |
You're gonna have your null hypothesis and your alternative hypothesis. Your null hypothesis is going to be a hypothesis about the population parameter that you care about, and it's going to assume kind of the status quo, no news here. And so the parameter that we care about is the mean amount of liquid in the bottles in the batch. So that's mu right over there. And what would be the assumption that that would be, the no news here? Well, it would be 500 milliliters. That's the target value. | Writing hypotheses for a significance test about a mean AP Statistics Khan Academy.mp3 |
So that's mu right over there. And what would be the assumption that that would be, the no news here? Well, it would be 500 milliliters. That's the target value. So it's reasonable to say, well, you know, the null is it's doing what it's supposed to, that where the actual mean for the batch is actually what the target needs to be, is actually 500 milliliters. Some of you might have said, hey, wait, didn't they say the amounts in the sample had a mean of 503 milliliters? Why isn't this 503? | Writing hypotheses for a significance test about a mean AP Statistics Khan Academy.mp3 |
That's the target value. So it's reasonable to say, well, you know, the null is it's doing what it's supposed to, that where the actual mean for the batch is actually what the target needs to be, is actually 500 milliliters. Some of you might have said, hey, wait, didn't they say the amounts in the sample had a mean of 503 milliliters? Why isn't this 503? Remember, your hypothesis is going to be about the population parameter, your assumption about the population parameter. This 503 milliliters right over here, this is a sample statistic. This is a sample mean that's trying to estimate this thing right over here. | Writing hypotheses for a significance test about a mean AP Statistics Khan Academy.mp3 |
Why isn't this 503? Remember, your hypothesis is going to be about the population parameter, your assumption about the population parameter. This 503 milliliters right over here, this is a sample statistic. This is a sample mean that's trying to estimate this thing right over here. When we do our significance test, we're going to incorporate this 533 milliliters. We're going to think about, well, what's the probability of getting a sample statistic, a sample mean, this far or further away from the assumed mean if we assume that the null hypothesis is true, and if that probability is below a threshold, our significance level, then we reject the null hypothesis, and it would suggest the alternative. But if we're just trying to generate or write a set of hypotheses, this would be our null hypothesis. | Writing hypotheses for a significance test about a mean AP Statistics Khan Academy.mp3 |
So this says a manager oversees 11 female employees and nine male employees. They need to pick three of these employees to go on a business trip, so the manager places all 20 names in a hat and chooses at random. Let x equal the number of female employees chosen. So they're going to do three trials, and on each of those trials, you could say success is if they pick a female employee, and then the random variable x is the number of females out of those three. Is x a binomial variable? Why or why not? So pause this video and see if you can work through this on your own. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
So they're going to do three trials, and on each of those trials, you could say success is if they pick a female employee, and then the random variable x is the number of females out of those three. Is x a binomial variable? Why or why not? So pause this video and see if you can work through this on your own. All right, now let's go through each choice. Choice A says each trial isn't being classified as a success or failure, so x is not a binomial variable. I disagree with this. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
So pause this video and see if you can work through this on your own. All right, now let's go through each choice. Choice A says each trial isn't being classified as a success or failure, so x is not a binomial variable. I disagree with this. Each trial is being classified as a success or failure. It's either going to be female or not, and since we're counting the number of female employees, if in a trial we pick a female, that would be a success, so each trial is being classified as a success or failure. So this one over here isn't true. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
I disagree with this. Each trial is being classified as a success or failure. It's either going to be female or not, and since we're counting the number of female employees, if in a trial we pick a female, that would be a success, so each trial is being classified as a success or failure. So this one over here isn't true. There is no fixed number of trials, so x is not a binomial variable. There is a fixed number of trials. They're doing three trials. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
So this one over here isn't true. There is no fixed number of trials, so x is not a binomial variable. There is a fixed number of trials. They're doing three trials. They're picking three hats, three names out of a hat. The trials are not independent, so x is not a binomial variable. So this is interesting. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
They're doing three trials. They're picking three hats, three names out of a hat. The trials are not independent, so x is not a binomial variable. So this is interesting. So for example, trial one, what's the probability of success? Well, there are 20 employees, 20 names in the hat, and 11 of the outcomes would be success, so you have 11 20th probability of success, but in trial two, what's the probability of success given success in, I'll say, trial one, T1? Well, if you succeeded in trial one, that means that there's now only 10 female names in the hat out of 19, and if you don't have success in trial one, then you will have, then it'll be 11 out of 19, so your probability does change based on previous outcomes, and so the trials are not independent, and so x is not a binomial variable. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
So this is interesting. So for example, trial one, what's the probability of success? Well, there are 20 employees, 20 names in the hat, and 11 of the outcomes would be success, so you have 11 20th probability of success, but in trial two, what's the probability of success given success in, I'll say, trial one, T1? Well, if you succeeded in trial one, that means that there's now only 10 female names in the hat out of 19, and if you don't have success in trial one, then you will have, then it'll be 11 out of 19, so your probability does change based on previous outcomes, and so the trials are not independent, and so x is not a binomial variable. So this one is true. The trials are not independent, so that violates that condition for being a binomial variable. In order to be a binomial variable, all your trials have to be independent of each other, and so we'd rule this last one out because this last one says that x has a binomial distribution, or it is or does meet all the conditions for being a binomial variable. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
Well, if you succeeded in trial one, that means that there's now only 10 female names in the hat out of 19, and if you don't have success in trial one, then you will have, then it'll be 11 out of 19, so your probability does change based on previous outcomes, and so the trials are not independent, and so x is not a binomial variable. So this one is true. The trials are not independent, so that violates that condition for being a binomial variable. In order to be a binomial variable, all your trials have to be independent of each other, and so we'd rule this last one out because this last one says that x has a binomial distribution, or it is or does meet all the conditions for being a binomial variable. Let's do another example. So here we have different scenarios, and I have the conditions for binomial variable written right over here, and so once again, pause the video and look at each of these scenarios for random variables, and look at these conditions, and think about whether these random variables are binomial or not. So let's look at the first one. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
In order to be a binomial variable, all your trials have to be independent of each other, and so we'd rule this last one out because this last one says that x has a binomial distribution, or it is or does meet all the conditions for being a binomial variable. Let's do another example. So here we have different scenarios, and I have the conditions for binomial variable written right over here, and so once again, pause the video and look at each of these scenarios for random variables, and look at these conditions, and think about whether these random variables are binomial or not. So let's look at the first one. In a game involving a standard deck of 52 playing cards, an individual randomly draws seven cards without replacement. Let y be equal to the number of aces drawn. Well, in our introductory video to binomial variables, we talked about if we're doing without replacement, your probability of getting an ace on a given trial, where trial is you're taking a card out of the deck, it's going to be dependent on whether you got aces in previous trials, because if you got an ace in a previous trial, well, that ace, then you're gonna have fewer aces in the decks. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
So let's look at the first one. In a game involving a standard deck of 52 playing cards, an individual randomly draws seven cards without replacement. Let y be equal to the number of aces drawn. Well, in our introductory video to binomial variables, we talked about if we're doing without replacement, your probability of getting an ace on a given trial, where trial is you're taking a card out of the deck, it's going to be dependent on whether you got aces in previous trials, because if you got an ace in a previous trial, well, that ace, then you're gonna have fewer aces in the decks. So the trials in this case are not independent. Not independent, not independent trials. Now, on the other hand, if on every trial you looked at whatever card you got and put it back in the deck, then they would be independent trials. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
Well, in our introductory video to binomial variables, we talked about if we're doing without replacement, your probability of getting an ace on a given trial, where trial is you're taking a card out of the deck, it's going to be dependent on whether you got aces in previous trials, because if you got an ace in a previous trial, well, that ace, then you're gonna have fewer aces in the decks. So the trials in this case are not independent. Not independent, not independent trials. Now, on the other hand, if on every trial you looked at whatever card you got and put it back in the deck, then they would be independent trials. The probability of getting an ace on each trial would be the same, but not when you have without replacement. So this is not binomial right over here, because you don't have independent trials. The second scenario. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
Now, on the other hand, if on every trial you looked at whatever card you got and put it back in the deck, then they would be independent trials. The probability of getting an ace on each trial would be the same, but not when you have without replacement. So this is not binomial right over here, because you don't have independent trials. The second scenario. 60% of a certain species of tomato live after transplanting from pot to garden. Eli transplants 16 of these tomato plants. Assume that the plants live independently of each other. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
The second scenario. 60% of a certain species of tomato live after transplanting from pot to garden. Eli transplants 16 of these tomato plants. Assume that the plants live independently of each other. So whether one plant lives isn't dependent on whether another plant lives. Let T equal the number of tomato plants that live. All right, so let's look at the conditions. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
Assume that the plants live independently of each other. So whether one plant lives isn't dependent on whether another plant lives. Let T equal the number of tomato plants that live. All right, so let's look at the conditions. The outcome of each trial can be classified as either success or failure. So each trial over here is one of the tomato plants, and we have 16 of those trials, and success is if the tomato plant lives, and failure is if it dies. So we have either success or failure. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
All right, so let's look at the conditions. The outcome of each trial can be classified as either success or failure. So each trial over here is one of the tomato plants, and we have 16 of those trials, and success is if the tomato plant lives, and failure is if it dies. So we have either success or failure. Each trial is independent of the others. They tell us the plants live independently of each other. So whether or not a neighboring plant lives or dies doesn't affect whether the plant next to it lives or dies. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
So we have either success or failure. Each trial is independent of the others. They tell us the plants live independently of each other. So whether or not a neighboring plant lives or dies doesn't affect whether the plant next to it lives or dies. So each trial is independent of the others. There is a fixed number of trials. Yes, we have 16 right over there. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
So whether or not a neighboring plant lives or dies doesn't affect whether the plant next to it lives or dies. So each trial is independent of the others. There is a fixed number of trials. Yes, we have 16 right over there. The probability P of success on each trial remains constant. Well, yeah, according to at least the scenario, they're saying that we have a 60% chance for each tomato plant, which is each trial. So it meets all of the conditions right over here. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
Yes, we have 16 right over there. The probability P of success on each trial remains constant. Well, yeah, according to at least the scenario, they're saying that we have a 60% chance for each tomato plant, which is each trial. So it meets all of the conditions right over here. So this one is binomial. Now let's look at this third scenario. In a game of luck, a turn consists of a player continuing to roll a pair of six-sided die until they roll a double, two of the same face values. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
So it meets all of the conditions right over here. So this one is binomial. Now let's look at this third scenario. In a game of luck, a turn consists of a player continuing to roll a pair of six-sided die until they roll a double, two of the same face values. Let x equal the number of rolls in one turn. So you're gonna keep rolling until they roll a double. Well, the thing that jumps out at me is that you don't have a fixed number of trials, not fixed number of trials. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
In a game of luck, a turn consists of a player continuing to roll a pair of six-sided die until they roll a double, two of the same face values. Let x equal the number of rolls in one turn. So you're gonna keep rolling until they roll a double. Well, the thing that jumps out at me is that you don't have a fixed number of trials, not fixed number of trials. You could say each trial, each roll is a trial. Success is getting a double, which has a fixed probability. Whether or not you get a double on each trial is gonna be independent of the previous roll. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
Well, the thing that jumps out at me is that you don't have a fixed number of trials, not fixed number of trials. You could say each trial, each roll is a trial. Success is getting a double, which has a fixed probability. Whether or not you get a double on each trial is gonna be independent of the previous roll. So it meets all the other constraints, but it does not meet that there's a fixed number of trials. You're gonna keep someone, there's some chance you might have to roll 20 times, or 200 times, or who knows however many times, until they roll a double. And so this violates there's a fixed number of trials. | Recognizing binomial variables Random variables AP Statistics Khan Academy.mp3 |
Maybe it's not your whole school, maybe it's just your grade. So there's 80 students in your population, and you want to get an estimate of the average height in your population. And you think it's too hard for you to go and measure the height of all 80 students, so you decide to find a simple, or take a simple random sample. You think it's reasonable for you to measure the heights of 30 of these students. And so what you want to do is randomly sample 30 of the 80 students, and take their average height, and say, well, that's probably a pretty good estimate for the population parameter, for the average height of the entire population. So once you decide to do this, you say, well, how do I select those 30 students, and how do I select it so that I feel good that it is actually random? And there's several ways that you could approach this. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
You think it's reasonable for you to measure the heights of 30 of these students. And so what you want to do is randomly sample 30 of the 80 students, and take their average height, and say, well, that's probably a pretty good estimate for the population parameter, for the average height of the entire population. So once you decide to do this, you say, well, how do I select those 30 students, and how do I select it so that I feel good that it is actually random? And there's several ways that you could approach this. One way to do it is associate every person in your school with a piece of paper, and put them all in a bowl, and then pick them out. So let's do that. So let's say this is alphabetically the first person in the school, they're on a slip of paper, then the next slip of paper gets the next person, and you're gonna go all the way down, so you're gonna have 80 pieces of paper. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
And there's several ways that you could approach this. One way to do it is associate every person in your school with a piece of paper, and put them all in a bowl, and then pick them out. So let's do that. So let's say this is alphabetically the first person in the school, they're on a slip of paper, then the next slip of paper gets the next person, and you're gonna go all the way down, so you're gonna have 80 pieces of paper. They all should be the same size. And then you throw them all, you throw them all into a bowl of some kind. And this seems like a very basic way of doing it, but it's actually a pretty effective way of getting a simple, of getting a simple random sample. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
So let's say this is alphabetically the first person in the school, they're on a slip of paper, then the next slip of paper gets the next person, and you're gonna go all the way down, so you're gonna have 80 pieces of paper. They all should be the same size. And then you throw them all, you throw them all into a bowl of some kind. And this seems like a very basic way of doing it, but it's actually a pretty effective way of getting a simple, of getting a simple random sample. So I'll try to draw a little, I don't know, it looks like a little fishbowl or something. All right, so that's our bowl. And so all the pieces of paper go in there. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
And this seems like a very basic way of doing it, but it's actually a pretty effective way of getting a simple, of getting a simple random sample. So I'll try to draw a little, I don't know, it looks like a little fishbowl or something. All right, so that's our bowl. And so all the pieces of paper go in there. And then you put a blindfold on someone, and they can't feel what names are there, and so they should pick out the first 30 without replacing them, because you obviously don't wanna pick the same, you don't wanna pick out the same name twice. And those 30 names that you pick, that would be your simple random sample, and then you could measure their heights to estimate the average height for the population. This would be a completely legitimate way of doing it. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
And so all the pieces of paper go in there. And then you put a blindfold on someone, and they can't feel what names are there, and so they should pick out the first 30 without replacing them, because you obviously don't wanna pick the same, you don't wanna pick out the same name twice. And those 30 names that you pick, that would be your simple random sample, and then you could measure their heights to estimate the average height for the population. This would be a completely legitimate way of doing it. Other ways that you could do it, if you have a computer or a calculator, you could use a random number generator. And the random functions on computer programming languages or on your calculator, they tend to be something, you know, some place you'll see something like a math.rand, short for random. You might see something like random. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
This would be a completely legitimate way of doing it. Other ways that you could do it, if you have a computer or a calculator, you could use a random number generator. And the random functions on computer programming languages or on your calculator, they tend to be something, you know, some place you'll see something like a math.rand, short for random. You might see something like random. You might see, you might see something like random. Without anything passed into it, it might give you a number between zero and one, or zero and 100. And you have to be very careful on how you use this to make sure that you have an even chance of picking certain numbers. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
You might see something like random. You might see, you might see something like random. Without anything passed into it, it might give you a number between zero and one, or zero and 100. And you have to be very careful on how you use this to make sure that you have an even chance of picking certain numbers. But what you would do in this situation, if you had access to some random number generator, and it could even pick out a random number between one and 80, including one and 80, is you would maybe line up all the students' names alphabetically. And so the first student alphabetically, assign the number zero, one. And you could just say one if you're using a random number generator, but I'll use two digits for it just because it'll be useful and consistent. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
And you have to be very careful on how you use this to make sure that you have an even chance of picking certain numbers. But what you would do in this situation, if you had access to some random number generator, and it could even pick out a random number between one and 80, including one and 80, is you would maybe line up all the students' names alphabetically. And so the first student alphabetically, assign the number zero, one. And you could just say one if you're using a random number generator, but I'll use two digits for it just because it'll be useful and consistent. And in a little bit, we'll use another technique where it's gonna be nice to be consistent with our number of digits. And so the next one, zero, two, and you go all the way to 79, and all the way to 80. And then you use your random number generator to keep generating numbers from one to 80. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
And you could just say one if you're using a random number generator, but I'll use two digits for it just because it'll be useful and consistent. And in a little bit, we'll use another technique where it's gonna be nice to be consistent with our number of digits. And so the next one, zero, two, and you go all the way to 79, and all the way to 80. And then you use your random number generator to keep generating numbers from one to 80. And as long as you don't get repeats, you pick the first 30 to be your actual random sample. Another related technique, which is a little bit more old school, but is definitely the way that it has been done in the past and even done now sometimes, is to use a random digit table. You still start with these number associations with each student in the class, and then you use a randomly generated list of numbers. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
And then you use your random number generator to keep generating numbers from one to 80. And as long as you don't get repeats, you pick the first 30 to be your actual random sample. Another related technique, which is a little bit more old school, but is definitely the way that it has been done in the past and even done now sometimes, is to use a random digit table. You still start with these number associations with each student in the class, and then you use a randomly generated list of numbers. And so let's say that's our randomly generated list of numbers, and it keeps going well beyond this. And you start at the beginning. And you say, okay, we're interested in getting 30 two-digit numbers from one to 80, including one in 80. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
You still start with these number associations with each student in the class, and then you use a randomly generated list of numbers. And so let's say that's our randomly generated list of numbers, and it keeps going well beyond this. And you start at the beginning. And you say, okay, we're interested in getting 30 two-digit numbers from one to 80, including one in 80. So one technique that you could use is you start it right at the beginning, and you could say, all right, this is a randomly generated list of numbers. So the first number here is 59. Is 59 between one and 80? | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
And you say, okay, we're interested in getting 30 two-digit numbers from one to 80, including one in 80. So one technique that you could use is you start it right at the beginning, and you could say, all right, this is a randomly generated list of numbers. So the first number here is 59. Is 59 between one and 80? Sure is. As long as we, you know, if this was a zero one, that would have worked. If this was an eight zero, that would have worked. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
Is 59 between one and 80? Sure is. As long as we, you know, if this was a zero one, that would have worked. If this was an eight zero, that would have worked. If this was a zero zero, it wouldn't have worked. If this was an eight one, it wouldn't have worked. But this would be our, this right over here, that would be our first name that we, you could imagine the same as picking that first name out of the hat, whoever's associated with number 59. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
If this was an eight zero, that would have worked. If this was a zero zero, it wouldn't have worked. If this was an eight one, it wouldn't have worked. But this would be our, this right over here, that would be our first name that we, you could imagine the same as picking that first name out of the hat, whoever's associated with number 59. Now, you would move on. You get the next two digits. The next two digits are 83. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
But this would be our, this right over here, that would be our first name that we, you could imagine the same as picking that first name out of the hat, whoever's associated with number 59. Now, you would move on. You get the next two digits. The next two digits are 83. They don't fall into our range from one to 80, so we're not going to use it. Then you look at the next two digits. So we get a five and a nine. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
The next two digits are 83. They don't fall into our range from one to 80, so we're not going to use it. Then you look at the next two digits. So we get a five and a nine. Well, that fits in our range, but we already picked 59. We already picked person 59, so we're not gonna pick 59 again. So we keep moving on. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
So we get a five and a nine. Well, that fits in our range, but we already picked 59. We already picked person 59, so we're not gonna pick 59 again. So we keep moving on. Then we get a 37. Well, that's in our range. We haven't picked that yet. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
So we keep moving on. Then we get a 37. Well, that's in our range. We haven't picked that yet. We do that. Then we get a zero zero. Once again, not in our range. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
We haven't picked that yet. We do that. Then we get a zero zero. Once again, not in our range. I think you see where this is going. 91, not in our range. 23, it's in our range, and we haven't picked it yet. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
Once again, not in our range. I think you see where this is going. 91, not in our range. 23, it's in our range, and we haven't picked it yet. So we're gonna pick the 23. I think you see where this is going. We're gonna keep going down this list in the way that I've just described until we get 30 of these. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
23, it's in our range, and we haven't picked it yet. So we're gonna pick the 23. I think you see where this is going. We're gonna keep going down this list in the way that I've just described until we get 30 of these. We've just gotten three. We just have to keep on going. And this isn't an exhaustive list of all of the different ways that you can get random numbers, but it starts to give you some techniques in your toolkit. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
We're gonna keep going down this list in the way that I've just described until we get 30 of these. We've just gotten three. We just have to keep on going. And this isn't an exhaustive list of all of the different ways that you can get random numbers, but it starts to give you some techniques in your toolkit. And you might say, oh, well, why don't I just randomly come up with some numbers in my head? And I would really suggest that you don't do that because humans are famously bad at being truly random. And you might wanna do something like even use something that you think is a random process, but you realize later that it wasn't as random as you thought. | Techniques for generating a simple random sample Study design AP Statistics Khan Academy.mp3 |
So what he does here is a simulation. It has a population that has a uniform distribution. So he says, I used a flat probabilistic distribution from 0 to 100 for my population. Then we start sampling from that population. We're going to use samples of size 50. And what we do is for each of those samples, we calculate the sample variance based on dividing by n by dividing by n minus 1 and n minus 2. And as we keep having more and more and more samples, we take the mean of the variances calculated in different ways. | Another simulation giving evidence that (n-1) gives us an unbiased estimate of variance.mp3 |
Then we start sampling from that population. We're going to use samples of size 50. And what we do is for each of those samples, we calculate the sample variance based on dividing by n by dividing by n minus 1 and n minus 2. And as we keep having more and more and more samples, we take the mean of the variances calculated in different ways. And we figure out what those means converge to. So that's a sample. Here's another sample. | Another simulation giving evidence that (n-1) gives us an unbiased estimate of variance.mp3 |
And as we keep having more and more and more samples, we take the mean of the variances calculated in different ways. And we figure out what those means converge to. So that's a sample. Here's another sample. Here's another sample. If I sample here, then now I'm adding a bunch. And I'm sampling continuously. | Another simulation giving evidence that (n-1) gives us an unbiased estimate of variance.mp3 |
Here's another sample. Here's another sample. If I sample here, then now I'm adding a bunch. And I'm sampling continuously. And you saw something very interesting happen. When I divide by n, I get my sample variance is still, even when I'm taking the mean of many, many, many, many sample variances that I've already taken, I'm still underestimating the true variance. When I divide by n minus 1, it looks like I'm getting a pretty good estimate. | Another simulation giving evidence that (n-1) gives us an unbiased estimate of variance.mp3 |
And I'm sampling continuously. And you saw something very interesting happen. When I divide by n, I get my sample variance is still, even when I'm taking the mean of many, many, many, many sample variances that I've already taken, I'm still underestimating the true variance. When I divide by n minus 1, it looks like I'm getting a pretty good estimate. The mean of all of my sample variances is really converged to the true variance. When I divided by n minus 2 just for kicks, it's pretty clear that I overestimated with my mean of my sample variances. I overestimated the true variance. | Another simulation giving evidence that (n-1) gives us an unbiased estimate of variance.mp3 |
When I divide by n minus 1, it looks like I'm getting a pretty good estimate. The mean of all of my sample variances is really converged to the true variance. When I divided by n minus 2 just for kicks, it's pretty clear that I overestimated with my mean of my sample variances. I overestimated the true variance. So this gives us a pretty good sense that n minus 1 is the right thing to do. Now, this is another way and another interesting way of visualizing it. In the horizontal axis right over here, we're comparing each plot as one of our samples. | Another simulation giving evidence that (n-1) gives us an unbiased estimate of variance.mp3 |