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The midrange, you could view it as the arithmetic mean or the average of your highest and lowest scores. So the midrange of midterm. So let's see. The highest midterm score, looking at the blue, the highest one is right here. So Jessica got 100 on the midterm. So that's your highest score. Your lowest score on the midterm looks like this one right over here. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
The highest midterm score, looking at the blue, the highest one is right here. So Jessica got 100 on the midterm. So that's your highest score. Your lowest score on the midterm looks like this one right over here. Daniel got a 60. And so the midrange is going to be the arithmetic mean of these two numbers. So you add 100 plus 60, divide by 2, you get 160 over 2, or 80. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
Your lowest score on the midterm looks like this one right over here. Daniel got a 60. And so the midrange is going to be the arithmetic mean of these two numbers. So you add 100 plus 60, divide by 2, you get 160 over 2, or 80. So this right over here is going to be 80. What was the average student score for the final exam? Well, for that, we just have to add up the scores on the final exams and then divide by the number of scores we have. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
So you add 100 plus 60, divide by 2, you get 160 over 2, or 80. So this right over here is going to be 80. What was the average student score for the final exam? Well, for that, we just have to add up the scores on the final exams and then divide by the number of scores we have. So we might be able to do that in our heads. Let's see. Well, let me just write it over here. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
Well, for that, we just have to add up the scores on the final exams and then divide by the number of scores we have. So we might be able to do that in our heads. Let's see. Well, let me just write it over here. So we have 100 plus 100 plus 100 plus 75 plus 80. So all of that divided by 5, that'll give us our average. If someone tells you average without giving more information, they're probably talking about the arithmetic mean. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
Well, let me just write it over here. So we have 100 plus 100 plus 100 plus 75 plus 80. So all of that divided by 5, that'll give us our average. If someone tells you average without giving more information, they're probably talking about the arithmetic mean. So this is going to be 300 plus another 155. So it's going to be 455 over 5. This is equal to 455 over 5. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
If someone tells you average without giving more information, they're probably talking about the arithmetic mean. So this is going to be 300 plus another 155. So it's going to be 455 over 5. This is equal to 455 over 5. And let's see. 5, this is going to be equal to 5 goes into 450 90 times and into 5 once. So this is going to be equal to 91. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
This is equal to 455 over 5. And let's see. 5, this is going to be equal to 5 goes into 450 90 times and into 5 once. So this is going to be equal to 91. So the average score for the final exam is a 91. What was the mode for the final exam scores? So the mode is the most common score. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
So this is going to be equal to 91. So the average score for the final exam is a 91. What was the mode for the final exam scores? So the mode is the most common score. So once again, we've listed all of them. And it's pretty clear that the most common score here is 100. 100 shows up three times, while a 75 only shows up once and an 80 only shows up once. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
So the mode is the most common score. So once again, we've listed all of them. And it's pretty clear that the most common score here is 100. 100 shows up three times, while a 75 only shows up once and an 80 only shows up once. So here, the most common score is 100. It shows up three times. What is the range of the midterm scores? | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
100 shows up three times, while a 75 only shows up once and an 80 only shows up once. So here, the most common score is 100. It shows up three times. What is the range of the midterm scores? So the range is literally the difference between the highest score and the lowest score. So the highest score, we already figured up. And this is for the midterm. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
What is the range of the midterm scores? So the range is literally the difference between the highest score and the lowest score. So the highest score, we already figured up. And this is for the midterm. So the highest score is 100. And we're going to subtract from that the lowest score, which was a 60. So the range is equal to 40. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
And this is for the midterm. So the highest score is 100. And we're going to subtract from that the lowest score, which was a 60. So the range is equal to 40. The midrange was the average of these two things. The range is just the difference between the two. So the range of the midterm scores is 40. | Reading bar charts putting it together with central tendency Pre-Algebra Khan Academy.mp3 |
Let C be the number of cake orders Liliana receives in a month until she first gets an order over the telephone. Assume the method of placing each cake order is independent. So C, if we assume a few things, is a classic geometric random variable. What tells us that? Well, the giveaway is that we're gonna keep doing these independent trials where the probability of success is constant and there's a clear success. A telephone order in this case is a success. The probability is 10% of it happening. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
What tells us that? Well, the giveaway is that we're gonna keep doing these independent trials where the probability of success is constant and there's a clear success. A telephone order in this case is a success. The probability is 10% of it happening. And we're gonna keep doing it until we get a success. So classic geometric random variable. Now they ask us, find the probability, the probability that it takes fewer than five orders for Liliana to get her first telephone order of the month. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
The probability is 10% of it happening. And we're gonna keep doing it until we get a success. So classic geometric random variable. Now they ask us, find the probability, the probability that it takes fewer than five orders for Liliana to get her first telephone order of the month. So it's really the probability that C is less than five. So like always, pause this video and have a go at it. And even if you struggle with it, that's even, that's better, your brain will be more primed for the actual solution that we can go through together. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
Now they ask us, find the probability, the probability that it takes fewer than five orders for Liliana to get her first telephone order of the month. So it's really the probability that C is less than five. So like always, pause this video and have a go at it. And even if you struggle with it, that's even, that's better, your brain will be more primed for the actual solution that we can go through together. All right. So I'm assuming you've had a go at it. So there's a couple of ways to approach it. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
And even if you struggle with it, that's even, that's better, your brain will be more primed for the actual solution that we can go through together. All right. So I'm assuming you've had a go at it. So there's a couple of ways to approach it. You could say, well look, this is just gonna be the probability that C is equal to one plus the probability that C is equal to two plus the probability that C is equal to three plus the probability that C is equal to four. And we can calculate it this way. What is the probability that C equals one? | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
So there's a couple of ways to approach it. You could say, well look, this is just gonna be the probability that C is equal to one plus the probability that C is equal to two plus the probability that C is equal to three plus the probability that C is equal to four. And we can calculate it this way. What is the probability that C equals one? Well, it's the probability that her very first order is a telephone order. And so we'll have 0.1. What's the probability that C equals two? | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
What is the probability that C equals one? Well, it's the probability that her very first order is a telephone order. And so we'll have 0.1. What's the probability that C equals two? Well, it's the probability that her first order is not a telephone order. So it's one minus 10%. There's a 90% chance it's not a telephone order. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
What's the probability that C equals two? Well, it's the probability that her first order is not a telephone order. So it's one minus 10%. There's a 90% chance it's not a telephone order. And that her second order is a telephone order. What about the probability C equals three? Well, her first two orders would not be telephone orders and her third order would be one. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
There's a 90% chance it's not a telephone order. And that her second order is a telephone order. What about the probability C equals three? Well, her first two orders would not be telephone orders and her third order would be one. And then C equals four? Well, her first three orders would not be telephone orders and her fourth one would. And we could get a calculator maybe and add all of these things up and we would actually get the answer. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
Well, her first two orders would not be telephone orders and her third order would be one. And then C equals four? Well, her first three orders would not be telephone orders and her fourth one would. And we could get a calculator maybe and add all of these things up and we would actually get the answer. But you're probably wondering, well, this is kinda hairy to type into a calculator. Maybe there is an easier way to tackle this. And indeed, there is. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
And we could get a calculator maybe and add all of these things up and we would actually get the answer. But you're probably wondering, well, this is kinda hairy to type into a calculator. Maybe there is an easier way to tackle this. And indeed, there is. So think about it. The probability that C is less than five, that's the same thing as one minus the probability that we don't have a telephone order in the first four. One minus the probability that no telephone order in first four orders. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
And indeed, there is. So think about it. The probability that C is less than five, that's the same thing as one minus the probability that we don't have a telephone order in the first four. One minus the probability that no telephone order in first four orders. So what's this? Well, because this is just saying, what's the probability we do have an order in the first four? So it's the same thing as one minus the probability that we don't have an order in the first four. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
One minus the probability that no telephone order in first four orders. So what's this? Well, because this is just saying, what's the probability we do have an order in the first four? So it's the same thing as one minus the probability that we don't have an order in the first four. And this is pretty straightforward to calculate. So this is going to be equal to one minus, and let me do this in another color so we know what I'm referring to. So what's the probability that we have no telephone orders in the first four orders? | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
So it's the same thing as one minus the probability that we don't have an order in the first four. And this is pretty straightforward to calculate. So this is going to be equal to one minus, and let me do this in another color so we know what I'm referring to. So what's the probability that we have no telephone orders in the first four orders? Well, the probability on a given order that you don't have a telephone order is 0.9. And then if that has to be true for the first four, well, it's gonna be 0.9 times 0.9 times 0.9 times 0.9, or 0.9 to the fourth power. So this is a lot easier to calculate. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
So what's the probability that we have no telephone orders in the first four orders? Well, the probability on a given order that you don't have a telephone order is 0.9. And then if that has to be true for the first four, well, it's gonna be 0.9 times 0.9 times 0.9 times 0.9, or 0.9 to the fourth power. So this is a lot easier to calculate. So let's do that. Let's get a calculator out. All right, so let me just take 0.9 to the fourth power is equal to, and then let me subtract that from one. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
So this is a lot easier to calculate. So let's do that. Let's get a calculator out. All right, so let me just take 0.9 to the fourth power is equal to, and then let me subtract that from one. So let me make that a negative, and then let me add one to it. And we get, there you go, 0.3439. So this is equal to 0.3439. | Cumulative geometric probability (less than a value) AP Statistics Khan Academy (2).mp3 |
The marching band is holding a raffle at a football game with two prizes. After the first ticket is pulled out and the winner determined, the ticket is taped to the prize. The next ticket is pulled out to determine the winner of the second prize. Are the two events independent? Explain. Now before we even think about this exact case, let's think about what it means for events to be independent. It means that the outcome of one event doesn't affect the outcome of the other event. | Independent or dependent probability event Precalculus Khan Academy.mp3 |
Are the two events independent? Explain. Now before we even think about this exact case, let's think about what it means for events to be independent. It means that the outcome of one event doesn't affect the outcome of the other event. Now in this situation, the first event, after the first ticket is pulled out and the winner determined, the ticket is taped to the prize. Then the next ticket is pulled out to determine the winner of the second prize. Now, the winner of the second prize, the possible winners, the possible outcomes for the second prize, is dependent on who was pulled out for the first prize. | Independent or dependent probability event Precalculus Khan Academy.mp3 |
It means that the outcome of one event doesn't affect the outcome of the other event. Now in this situation, the first event, after the first ticket is pulled out and the winner determined, the ticket is taped to the prize. Then the next ticket is pulled out to determine the winner of the second prize. Now, the winner of the second prize, the possible winners, the possible outcomes for the second prize, is dependent on who was pulled out for the first prize. You can imagine, if there's three tickets, let's say there's tickets A, B, and C in the bag, and for the first prize, they pull out ticket A. That's for the first prize. Now, when we think about who could be pulled out for the second prize, it's only going to be tickets B or C. Now, the first prize could have gone the other way. | Independent or dependent probability event Precalculus Khan Academy.mp3 |
Now, the winner of the second prize, the possible winners, the possible outcomes for the second prize, is dependent on who was pulled out for the first prize. You can imagine, if there's three tickets, let's say there's tickets A, B, and C in the bag, and for the first prize, they pull out ticket A. That's for the first prize. Now, when we think about who could be pulled out for the second prize, it's only going to be tickets B or C. Now, the first prize could have gone the other way. It could have been A, B, and C. The first prize could have gone to ticket B. And then the possible outcomes for the second prize would be A or C. So the possible outcomes for the second event, for the second prize, are completely dependent, are completely dependent on what happened, or what ticket was pulled out for the first prize. So these are not independent events. | Independent or dependent probability event Precalculus Khan Academy.mp3 |
Now, when we think about who could be pulled out for the second prize, it's only going to be tickets B or C. Now, the first prize could have gone the other way. It could have been A, B, and C. The first prize could have gone to ticket B. And then the possible outcomes for the second prize would be A or C. So the possible outcomes for the second event, for the second prize, are completely dependent, are completely dependent on what happened, or what ticket was pulled out for the first prize. So these are not independent events. The second event, the outcomes for it, are dependent on what happened in the first event. So they are not independent. The way that we could have made them independent is after the first ticket was pulled out, if they just wrote down the name or something, and then put that ticket back in. | Independent or dependent probability event Precalculus Khan Academy.mp3 |
So these are not independent events. The second event, the outcomes for it, are dependent on what happened in the first event. So they are not independent. The way that we could have made them independent is after the first ticket was pulled out, if they just wrote down the name or something, and then put that ticket back in. Instead, they taped it to the prize. But if they put that ticket back in, then the second prize, it would have still had all the tickets there. It wouldn't have mattered who was picked out in the first time, because their name was just written down, but their ticket was put back in, and then you would have been independent. | Independent or dependent probability event Precalculus Khan Academy.mp3 |
Let's do a couple of exercises from our probability one module. So we have a bag with nine red marbles, two blue marbles, and three green marbles in it. What is the probability of randomly selecting a non-blue marble from the bag? So let's draw this bag here. So that's my bag, and we're going to assume that it's a transparent bag. That looks like a vase. But we have nine red marbles, so let me draw nine red marbles. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So let's draw this bag here. So that's my bag, and we're going to assume that it's a transparent bag. That looks like a vase. But we have nine red marbles, so let me draw nine red marbles. One, two, three, four, five, six, seven, eight, nine red marbles. So that looks kind of orange-ish, but it does the job. Two blue marbles. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
But we have nine red marbles, so let me draw nine red marbles. One, two, three, four, five, six, seven, eight, nine red marbles. So that looks kind of orange-ish, but it does the job. Two blue marbles. So we have one blue marble, two blue marbles. And then we have three green marbles. Let me draw those three. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Two blue marbles. So we have one blue marble, two blue marbles. And then we have three green marbles. Let me draw those three. So one, two, three. What is the probability of randomly selecting a non-blue marble from the bag? So maybe we mix them all up, and we have an equal probability of selecting any one of these. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Let me draw those three. So one, two, three. What is the probability of randomly selecting a non-blue marble from the bag? So maybe we mix them all up, and we have an equal probability of selecting any one of these. And the way you just think about it is, is what fraction of all of the possible events meet our constraints? So let's just think about all of the possible events first. How many different possible marbles can we take out? | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So maybe we mix them all up, and we have an equal probability of selecting any one of these. And the way you just think about it is, is what fraction of all of the possible events meet our constraints? So let's just think about all of the possible events first. How many different possible marbles can we take out? Well, that's just the total number of marbles there are. So there are one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen possible marbles. So this is the number of possibilities. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
How many different possible marbles can we take out? Well, that's just the total number of marbles there are. So there are one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen possible marbles. So this is the number of possibilities. And then we just have to think, what fraction of those possibilities meet our constraints? And the other way you could have gotten fourteen is just taking nine plus two plus three. So what number of those possibilities meet our constraints? | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So this is the number of possibilities. And then we just have to think, what fraction of those possibilities meet our constraints? And the other way you could have gotten fourteen is just taking nine plus two plus three. So what number of those possibilities meet our constraints? And remember, our constraint is selecting a non-blue marble from the bag. Another way to think about it is a red or a green marble, because the only non-blue ones, the only other two colors we have are red and green. So how many non-blue marbles are there? | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So what number of those possibilities meet our constraints? And remember, our constraint is selecting a non-blue marble from the bag. Another way to think about it is a red or a green marble, because the only non-blue ones, the only other two colors we have are red and green. So how many non-blue marbles are there? Well, there's a couple of ways to think about it. You could say there's fourteen total marbles. Two are blue, so they're going to be fourteen minus two, which is twelve non-blue marbles. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So how many non-blue marbles are there? Well, there's a couple of ways to think about it. You could say there's fourteen total marbles. Two are blue, so they're going to be fourteen minus two, which is twelve non-blue marbles. Or you could just count them. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So there are twelve non-blue marbles. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Two are blue, so they're going to be fourteen minus two, which is twelve non-blue marbles. Or you could just count them. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So there are twelve non-blue marbles. So that's the number of non-blue. So these are the possibilities that meet our constraints over all of the possibilities. And then if we want to, this isn't in simplified form right here, since both twelve and fourteen are divisible by two. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So there are twelve non-blue marbles. So that's the number of non-blue. So these are the possibilities that meet our constraints over all of the possibilities. And then if we want to, this isn't in simplified form right here, since both twelve and fourteen are divisible by two. So let's divide both the numerator and the denominator by two, and you get six over seven. So we have a six-seventh chance of selecting a non-blue marble from the bag. Let's do another one. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
And then if we want to, this isn't in simplified form right here, since both twelve and fourteen are divisible by two. So let's divide both the numerator and the denominator by two, and you get six over seven. So we have a six-seventh chance of selecting a non-blue marble from the bag. Let's do another one. If a number is randomly chosen from the following list, what is the probability that the number is a multiple of five? So once again, we want to find the fraction of the total possibilities that meet our constraint. And our constraint is being a multiple of five. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Let's do another one. If a number is randomly chosen from the following list, what is the probability that the number is a multiple of five? So once again, we want to find the fraction of the total possibilities that meet our constraint. And our constraint is being a multiple of five. So how many total possibilities are there? Let's think about that. Total possibilities. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
And our constraint is being a multiple of five. So how many total possibilities are there? Let's think about that. Total possibilities. How many do we have? Well, that's just the total number of numbers we have to pick from. So one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Total possibilities. How many do we have? Well, that's just the total number of numbers we have to pick from. So one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So there are twelve possibilities. We have an equal chance of picking any one of these twelve. Now, which of these twelve are a multiple of five? | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. So there are twelve possibilities. We have an equal chance of picking any one of these twelve. Now, which of these twelve are a multiple of five? So let's do this in a different color. So let me pick out the multiples of five. Thirty-two is not a multiple of five. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Now, which of these twelve are a multiple of five? So let's do this in a different color. So let me pick out the multiples of five. Thirty-two is not a multiple of five. Forty-nine is not a multiple of five. Fifty-five is a multiple of five. Really, we're just looking for the numbers that, in the ones place, either have a five or a zero. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Thirty-two is not a multiple of five. Forty-nine is not a multiple of five. Fifty-five is a multiple of five. Really, we're just looking for the numbers that, in the ones place, either have a five or a zero. Fifty-five is a multiple of five. Thirty is a multiple of five. That's six times five. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Really, we're just looking for the numbers that, in the ones place, either have a five or a zero. Fifty-five is a multiple of five. Thirty is a multiple of five. That's six times five. Fifty-five is eleven times five. Not fifty-six, not twenty-eight. This is clearly five times ten. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
That's six times five. Fifty-five is eleven times five. Not fifty-six, not twenty-eight. This is clearly five times ten. This is eight times five. This is the same number again. Also eight times five. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
This is clearly five times ten. This is eight times five. This is the same number again. Also eight times five. So all of these are multiples of five. Forty-five, that's a nine times five. Three is not a multiple of five. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Also eight times five. So all of these are multiples of five. Forty-five, that's a nine times five. Three is not a multiple of five. Twenty-five, clearly five times five. So I've circled all of the multiples of five. So of all the possibilities, the ones that meet our constraint of being a multiple of five, there are one, two, three, four, five, six, seven possibilities. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Three is not a multiple of five. Twenty-five, clearly five times five. So I've circled all of the multiples of five. So of all the possibilities, the ones that meet our constraint of being a multiple of five, there are one, two, three, four, five, six, seven possibilities. So seven meet our constraint. So in this example, the probability of selecting a number that is a multiple of five is 7 twelfths. Let's do another one. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So of all the possibilities, the ones that meet our constraint of being a multiple of five, there are one, two, three, four, five, six, seven possibilities. So seven meet our constraint. So in this example, the probability of selecting a number that is a multiple of five is 7 twelfths. Let's do another one. The circumference of a circle is 36 pi. Let's draw this circle. The circumference of a circle is 36 pi. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Let's do another one. The circumference of a circle is 36 pi. Let's draw this circle. The circumference of a circle is 36 pi. So let's say the circle looks something like that. And the circumference, we have to be careful here. They're giving us interesting. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
The circumference of a circle is 36 pi. So let's say the circle looks something like that. And the circumference, we have to be careful here. They're giving us interesting. So the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has this guy right over here has an area of 16 pi. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
They're giving us interesting. So the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has this guy right over here has an area of 16 pi. A point is selected at random from inside the larger circle. So we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So inside the bigger circle, we have a smaller circle that has this guy right over here has an area of 16 pi. A point is selected at random from inside the larger circle. So we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? So here's a little bit interesting because you actually have an infinite number of points in both of these circles. It's not kind of separate balls or marbles like we saw in the first example or separate numbers. There's actually an infinite number of points you can pick here. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
What is the probability that the point also lies in this smaller circle? So here's a little bit interesting because you actually have an infinite number of points in both of these circles. It's not kind of separate balls or marbles like we saw in the first example or separate numbers. There's actually an infinite number of points you can pick here. So when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we were to pick a point from this larger circle, the probability that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we really just have to figure out the areas for both of them. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
There's actually an infinite number of points you can pick here. So when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we were to pick a point from this larger circle, the probability that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we really just have to figure out the areas for both of them. It's really just going to be the ratios. So let's think about that. So there's a temptation to just use this 36 pi up here. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
I know that might sound confusing, but we really just have to figure out the areas for both of them. It's really just going to be the ratios. So let's think about that. So there's a temptation to just use this 36 pi up here. We have to remember this was the circumference and we need to figure out the area of both of these circles. And so for area, we need to know the radius because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So there's a temptation to just use this 36 pi up here. We have to remember this was the circumference and we need to figure out the area of both of these circles. And so for area, we need to know the radius because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle. Or if you say 36 pi, which we're told is a circumference, is equal to 2 times pi times the radius. We can divide both sides by 2 pi. And on the left-hand side, 36 divided by 2 is 18. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle. Or if you say 36 pi, which we're told is a circumference, is equal to 2 times pi times the radius. We can divide both sides by 2 pi. And on the left-hand side, 36 divided by 2 is 18. The pi's cancel out. We get our radius as being equal to 18 for this larger circle. This larger circle has a radius of 18. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
And on the left-hand side, 36 divided by 2 is 18. The pi's cancel out. We get our radius as being equal to 18 for this larger circle. This larger circle has a radius of 18. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. Now let's figure out what 18 squared is. 18 times 18. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
This larger circle has a radius of 18. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. Now let's figure out what 18 squared is. 18 times 18. 8 times 8 is 64. 8 times 1 is 8. Plus 6 is 14. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
18 times 18. 8 times 8 is 64. 8 times 1 is 8. Plus 6 is 14. And then we have, we put that 0 there because we're now in the tens place. 1 times 8 is 8. 1 times 1 is 1. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
Plus 6 is 14. And then we have, we put that 0 there because we're now in the tens place. 1 times 8 is 8. 1 times 1 is 1. And really this is a 10 times a 10. That's why it gives us 100. But anyway, 4 plus 0 is a 4. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
1 times 1 is 1. And really this is a 10 times a 10. That's why it gives us 100. But anyway, 4 plus 0 is a 4. 4 plus 8 is a 12. 1 plus 1 plus 1 is a 3. So it's 324. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
But anyway, 4 plus 0 is a 4. 4 plus 8 is a 12. 1 plus 1 plus 1 is a 3. So it's 324. So the area here is equal to pi times 324, or we could say 324 pi. So the area of the entire larger circle, the part that I've shaded in yellow, including what's kind of under this orange circle, if you want to view it that way, this area right over here is equal to 324 pi. So the probability that a point that we select from this larger circle is also in the smaller circle is really just a percentage of the larger circle that is the smaller circle. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So it's 324. So the area here is equal to pi times 324, or we could say 324 pi. So the area of the entire larger circle, the part that I've shaded in yellow, including what's kind of under this orange circle, if you want to view it that way, this area right over here is equal to 324 pi. So the probability that a point that we select from this larger circle is also in the smaller circle is really just a percentage of the larger circle that is the smaller circle. So our probability, I'll just write it like this, the probability that the point also lies in the smaller circle, so all of that stuff I'll put in, the probability of that is going to be equal to the percentage of this larger circle that is the smaller one. So that's going to be, or we could say the fraction of the larger circle's area that is the smaller circle's area. So it's going to be 16 pi over 324 pi. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So the probability that a point that we select from this larger circle is also in the smaller circle is really just a percentage of the larger circle that is the smaller circle. So our probability, I'll just write it like this, the probability that the point also lies in the smaller circle, so all of that stuff I'll put in, the probability of that is going to be equal to the percentage of this larger circle that is the smaller one. So that's going to be, or we could say the fraction of the larger circle's area that is the smaller circle's area. So it's going to be 16 pi over 324 pi. And the pi's cancel out. And let's see, it looks like both of them are divisible by 4. If we divide the numerator by 4, we get 4. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
So it's going to be 16 pi over 324 pi. And the pi's cancel out. And let's see, it looks like both of them are divisible by 4. If we divide the numerator by 4, we get 4. If we divide the denominator by 4, what do we get? 4 goes into 320, 80 times, it goes into 4 once, so we get an 81. So the probability, so I didn't even draw this to scale, this area is actually much smaller when you do it to scale. | Finding probability example 2 Probability and Statistics Khan Academy.mp3 |
He's going to choose a volunteer, and he wants each kid to have the same chance of getting chosen. Fair enough. Now we have three methods by which he can do it, and let's just think about whether each of these methods are fair, where each kid does have the same chance of getting chosen, and if they're not, if they don't lead to each kid having the same chance of getting chosen, think about why that is the case. So method one, the magician starts with the birthday boy and moves clockwise, passing out 100 pieces of paper numbered one through 100. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer one through 100 and chooses the volunteer with that number. So let's just think about what's happening. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So method one, the magician starts with the birthday boy and moves clockwise, passing out 100 pieces of paper numbered one through 100. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer one through 100 and chooses the volunteer with that number. So let's just think about what's happening. So there's 15 kids in a circle. So there's one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. I planned that out amazingly well. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So let's just think about what's happening. So there's 15 kids in a circle. So there's one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. I planned that out amazingly well. I didn't think I would be able to fit exactly 15, but it worked out. So 15 kids in a circle, and then he's gonna hand out pieces of paper. So he's going to, let's give one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
I planned that out amazingly well. I didn't think I would be able to fit exactly 15, but it worked out. So 15 kids in a circle, and then he's gonna hand out pieces of paper. So he's going to, let's give one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. And now this person's gonna get pieces of paper one and 16. Now this person's gonna get two and 17. You're gonna keep going around and around and around until all 100 pieces of paper are going to get distributed. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So he's going to, let's give one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. And now this person's gonna get pieces of paper one and 16. Now this person's gonna get two and 17. You're gonna keep going around and around and around until all 100 pieces of paper are going to get distributed. Now, something to think about is whether every child here is going to get the same number of pieces of paper. And I encourage you to pause this video and think about that. If we just keep cycling around all the way to 100, does each child get the same number of pieces of paper? | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
You're gonna keep going around and around and around until all 100 pieces of paper are going to get distributed. Now, something to think about is whether every child here is going to get the same number of pieces of paper. And I encourage you to pause this video and think about that. If we just keep cycling around all the way to 100, does each child get the same number of pieces of paper? Well, just think about it. In order to get the same number of pieces of paper, 100 has to be divisible by 15. And we know 100 isn't divisible by 15. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
If we just keep cycling around all the way to 100, does each child get the same number of pieces of paper? Well, just think about it. In order to get the same number of pieces of paper, 100 has to be divisible by 15. And we know 100 isn't divisible by 15. 15 goes into 100 six times. Six times 15 is 90. And you have a remainder of 10. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
And we know 100 isn't divisible by 15. 15 goes into 100 six times. Six times 15 is 90. And you have a remainder of 10. So what's going to happen is all 15 kids are going to get six pieces of paper. And then another 10 of the 15 are going to get a seventh piece of paper. So they're not all getting an equal number of pieces of paper. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
And you have a remainder of 10. So what's going to happen is all 15 kids are going to get six pieces of paper. And then another 10 of the 15 are going to get a seventh piece of paper. So they're not all getting an equal number of pieces of paper. So even though he's randomly picking an integer between one and 100, some of the students are going to have a higher chance than the other ones. The 10 that have seven pieces of paper are going to have a higher chance than the other five who only have six pieces, who only have six pieces of paper. And so I would say method one is not fair. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So they're not all getting an equal number of pieces of paper. So even though he's randomly picking an integer between one and 100, some of the students are going to have a higher chance than the other ones. The 10 that have seven pieces of paper are going to have a higher chance than the other five who only have six pieces, who only have six pieces of paper. And so I would say method one is not fair. Let me write this down. Not fair. Sometimes life isn't fair. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
And so I would say method one is not fair. Let me write this down. Not fair. Sometimes life isn't fair. But in this case, it's not fair. Where we define fair is the same chance of getting chosen. And that's because they all have different numbers of pieces of paper. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Sometimes life isn't fair. But in this case, it's not fair. Where we define fair is the same chance of getting chosen. And that's because they all have different numbers of pieces of paper. All of the students are not equally likely to get picked. Let's look at method two. The magician starts with the birthday boy and moves counterclockwise, passing up 75 pieces of paper, numbered one through 75. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
And that's because they all have different numbers of pieces of paper. All of the students are not equally likely to get picked. Let's look at method two. The magician starts with the birthday boy and moves counterclockwise, passing up 75 pieces of paper, numbered one through 75. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer between one through 75 and chooses to volunteer with that number. So I encourage you to pause this video and think about whether this one, method two, whether that one is fair. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
The magician starts with the birthday boy and moves counterclockwise, passing up 75 pieces of paper, numbered one through 75. He cycles around the circle until all the pieces are distributed. He then uses a random number generator to pick an integer between one through 75 and chooses to volunteer with that number. So I encourage you to pause this video and think about whether this one, method two, whether that one is fair. Well, method two is the same as method one, except for instead of using 100 pieces of paper, we're using 75 pieces of paper. And so we have to think about is 75 divisible by 15? And 75 is. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So I encourage you to pause this video and think about whether this one, method two, whether that one is fair. Well, method two is the same as method one, except for instead of using 100 pieces of paper, we're using 75 pieces of paper. And so we have to think about is 75 divisible by 15? And 75 is. 5 times 15 is 75. So in this situation, each student is going to get five pieces of paper. Each gets five pieces of paper. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
And 75 is. 5 times 15 is 75. So in this situation, each student is going to get five pieces of paper. Each gets five pieces of paper. So they all have an equally likely chance of getting picked. And then he's using a random number generator to pick them. So they all have an equally likely chance. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Each gets five pieces of paper. So they all have an equally likely chance of getting picked. And then he's using a random number generator to pick them. So they all have an equally likely chance. I would say method two is indeed fair. They all have the same chance of getting chosen. Now let's think about method three. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So they all have an equally likely chance. I would say method two is indeed fair. They all have the same chance of getting chosen. Now let's think about method three. The magician starts with the birthday boy and moves clockwise, passing out 30 pieces of paper numbered one through 30. So they're all going to get the same number of pieces of paper. They're all going to get two pieces of paper each. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
Now let's think about method three. The magician starts with the birthday boy and moves clockwise, passing out 30 pieces of paper numbered one through 30. So they're all going to get the same number of pieces of paper. They're all going to get two pieces of paper each. 15 children getting two pieces of paper each would be 30 pieces of paper. So that looks reasonable so far. He cycles around the circle until all the pieces are distributed. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
They're all going to get two pieces of paper each. 15 children getting two pieces of paper each would be 30 pieces of paper. So that looks reasonable so far. He cycles around the circle until all the pieces are distributed. So everyone gets two pieces. He gives number one to the birthday boy, number two to the next kid, and so on. So that all seems reasonable, kind of consistent with method two, except now instead of 75, it's 30. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
He cycles around the circle until all the pieces are distributed. So everyone gets two pieces. He gives number one to the birthday boy, number two to the next kid, and so on. So that all seems reasonable, kind of consistent with method two, except now instead of 75, it's 30. And obviously, 75 was overkill. Even here, this is overkill. He just really needs 15 pieces of paper. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
So that all seems reasonable, kind of consistent with method two, except now instead of 75, it's 30. And obviously, 75 was overkill. Even here, this is overkill. He just really needs 15 pieces of paper. And he then counts the number of windows in the room and chooses the volunteer with that number. So the question here is, is the number of windows in the room, is it random, and is it evenly distributed? So maybe you could make a case that depending on what building it's in and someone's house, it's somewhat random on how many windows that house happens to have, the house that's happening to host the birthday party. | Picking fairly Probability and combinatorics Probability and Statistics Khan Academy.mp3 |
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