Simple probability: non-blue marble

Let's do a couple of excercises from our probability one module. 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 lets draw this bag here. So that's my bag. We're going to assume that it's a transparent bag. That looks like a vase. We have nine red marbles. So we draw nine red marbles. One, two, three, four, five, six, seven, eight, nine red marbles. That looks some kind of orangesh but does the job. Two blue marbles. So we have one blue marble. Two blue marbles. And then we have three green marbles. Three green marbles. Let me draw those three...One, two, three. What is the probability of randomly selecting a non-blue marble from the bag? 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: What fraction of all of the possible events meet our constraints. So lets just think about all of the possible events first. How many different possible marbles can we take out? Thats just the total number of marbles their are. 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? The other way you could have gotten fourteen was by taking 9 + 2 + 3. So what number of the possiblities meet our constraints. And remember our constraint is selecting a non-blue marble from the bag. Another way of thinking about it is a red or a green marble. So how many non-blue marbles are there? Well there's a couple of ways of think about it. You could say there is 14 total marbles: two are blue, so there are going to be 14 minus 2 which is 12 non-blue marbles. Or, you could just count them: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. 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 lowest simplified... this isn't in simplified form right here, since both 12 and 14 are divisible by 2- lets divide both the numerator and the denominator by 2 and you get 6 over 7. 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. So how many total possibilities are there? Let's think about that. Total possibilities... How many do we have? That's just the total number of numbers we have to pick from. So that's just 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 possibilites. We have an equal chance of picking anyone of these 12. Now, which of these twelve are a multiple of five? So let's we do this in a different color, so let me pick out the multiples of five. 32 is not a multiple of five, 49 is not a multiple of five, 55 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. 55 is a multiple of five, 30 is a multiple of 5, that's 6 times 5, 55 is 11 times 5, not 56, not 28, This is clearly 5 times 10, this is 8 times 5, this is the same number again, also 8 times 5, so all of these are multiples of 5, 45, that's a 9 times 5, 3 not a multiple of 5, 25 clearly 5 times 5. So I circled all the multiples of 5. So, of all the possibilities, the ones that meet our constraint of being a multiple of five, there are : 1, 2, 3, 4, 5, 6, 7 possibilities. So 7 meet our constraint. So, in this example, the probability of selecting a number that is a multiple of 5, is seven twelves. Let's do another one. The circumference of a circle is 36 pi. Let's draw this. So the circle looks like... I can draw a neater circle than that. So let's say the circle loks something like that. And its circumference, we have to be carefull here, they're giving us interesting... so circumference ... 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 that has an area of 16 pi. A point is selected at random from inside the larger circle, so we're gonna 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, 'cause you actually have an infinite number of points in both of these circles, because it's ... 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 could pick here, and so when we talk about the probability that the point also lies in this 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 the point is also ... the probability that if we'll pick a point from this larger circle, the probability that'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, and it's just really going to be the ratio. So let's think about that: So there's a temptation to just use this 36 pi up here, but 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 'cause area's pi*R squared, so we can figure out the radius from the circumference by saying, well circumference is equal to two times pi times the radius of the circle, or if you say 36 pi, which we were told this is the circumference, is equal to two times pi times the radius. We can divide both sides by 2pi , and on the left hand side, 36 divided by 2 is 18, the pi's cancel out... we get our radius is being equal to 18, for this larger circle. This larger circle has a radius of 18. So if we want to know it's area, it's area it's going to be pi*R sqared. Which is equal to pi times 18 square. Let's figure out what 18 square is. 18 times 18: 8 times 8 is 64, 8 times 1 is 8, plus 6 is 14, and then we have that zero there 'cause we're not in the ten's place, 1 times 8 is 8, 1 times 1 is 1, and really this is a 10 times a 10, that's why gives us 100. Anyway... 4 plus 0 is a 4, 4 plus 8 is a 12, and then 1 plus 1 plus 1 is 3, so thet'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 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 324pi. 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 of ... that the point is also lies in the smaller circle .. so all of that stuff I'll put it ... the probability of that is going to be equal to the percentage of this larger circle that it is the smaller one so that's going to be... well... the fraction of the larger circle's area that is the smaller circle's area. So it's going to be 16pi over 324pi. And the pi's cancel out and 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 in 320, 80 times and goes into 4 once, so we get an 81. So the probability ... I didn't even draw this to scale, this area is actually much smaller when you do it to scale... The probability that you... if you are to randomly select a point from a larger circle that also lies in the smaller one is the ratio of the areas the ratio of the smaller circle to the larger one and that is four eighty ones, I guess is the best way to say it.