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Khan Academy Presents: SAT Prep: Problems 5-6 starting on page 408

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We are now on problem number five on page 408. And they have this little drawing here and it says on the disc shown above, a player spins the arrow—arrow here or this little blue arrow. A player spins the arrow twice. The fraction a/b is formed where a is a number of the sector where the arrow stops after the first spin. And b is the number of the sector where the arrow stops at the second spin. On every spin, each of the numbered sectors has an equal probability of being the sector on which the arrow stops. What is the probability that the fraction a/b is greater than one? So, they want to know the probability that a /b>1. Well, the first thing you had to realize is this is just a random number generator. The probability that any number between one and six is equal so this could just be a roll of six-sided die and say, “Well, if I roll a six-sided die twice, what are the odds? What’s the probability that the first roll is going to be larger than the second roll, right? Because in order for this fraction to be greater than one, the numerator has to be greater than the denominator” or you can just take a/b>1 and multiply it both sides by d and that’s the same thing as the probability that a>b. So, what we can do is we can set up a little table and we could say, “Well, what’s the probability of getting each of these numbers on the first row? And then given the probability or given that you got a two on the first roll, what are the odds that the second roll is going to be less than the first roll? So let’s say, roll or spin. Let’s call this spin one, two, three, four, five, six, and what’s the probability of getting each of these spins? Well, the probability on the first time—probability of spin so we call it spin one. I’m just making up notation as I go but hopefully this makes sense to you. So, the probability of getting anyone of these is 1/6 and they all have an equal probability of happening, 1/6. We have to say now, given each of these, what’s the probability that you spin two, your second spin is going to be less than, right? The second spin has to be less than the first spin. This is the second spin, probability of a. That will probably make more sense to you. And then, we want to know the probability that the second spin b is less than a. Well, if I got a one the first time, what’s the probability that I get a number less than one out of this? Well, is there any number less than one? There’s one but that’s not even less than, that’s equal to one. So, there’s a zero chance that I get a number less than one because there is no number less than one that I can get. If I got a two, well, there’s only one number less than one out of the six, that’s one. So, there’s a 1/6 chance. If I got a three on the first spin, I can only get a one or two on the second spin so that’s a 2/6 chance. If I got a four, I can get a one, two, or three. That’s a 3/6. You see the pattern here, there’s a 4/6 chance of getting something less than five. And there’s a 5/6 chance in getting something less than six. And so, what’s the combined probability now for each of distant areas? So, what is the probability that I get a one and then I get something less than one? Well, you just have to multiply these probabilities. So, that’s a zero. What’s the probability that I get a two and then something less than two? It’s 1/6*1/6 which is 1/36. Probability I get a three and then something less than three, multiply them, it’s 2/36. I think you see the pattern here again, 3/36, 4/36, then 5/36. 5/36 is the probability I get a six and then something less than a six. So in general, the probability that I get at my first spin is greater than my second spin is going to be the sum of these probabilities because it can be anyone of these scenarios would work, would satisfy a being greater than b. So, I’ll just add up these probabilities. So, let’s see. It’s like the denominator for all of them is 36. So, it’s essentially over 36, 1+2+3+4+5, let me add carefully because this is where I historically messed up my problems, 1+2=3, 3+3=6, 6+4=10, 10+5=15, so that equals 15/36 and that is choice a. Move on to the next problem, image, clear image, image invert colors. Which of the fallacy—we’re on problem number six, which of the following table’s shows a relationship in which w is proportional to x? So, let’s just write them all out. So, choice a—so if you have wx so they have my one, three, two, four, three, five so w is proportional to x. Well, that also means that x is going to be proportional to w. So, in this case, x is three times of—here we have times three is equal to three, here we have times two and here we have times five-thirds. So, we’re multiplying by different factor every time. So, w is not proportional to x and x is not proportional to w. I’m just trying to figure out what do I have to multiply w by to get x or whether I have to multiply x by to get w? And they all have to be the same in order for these two variables to be proportional. So, choice A is not right. Choice B. WX, 3/9, so I multiplied 3/9, 4/16, and 5/25. It looks like in every scenario here, we squared w to get x or w is a square root of x, but they are not proportional because here we multiply it by three. Here, we multiply it by four and here we multiply it by five. So once again, it’s not going to be choice B. C, I will change colors for variety. C and wx, 5, 10 so I multiply it by two, 6 to 18, I multiply it by 3 I already know this is wrong. I multiply it by two different factors. It’s not choice C. Choice D. W, they probably did at one of the later choices just to make sure we do a lot of work. Choice D is seven to 21. I multiply it by three to do that. 8 to 24 times three, it looks good so far. 9 to 27 multiply it by three. So in every case, I multiply it by the same factor so w is proportional to x or proportional to x and x is proportionate to w. So, D is our choice. I will see you in the next video. And we don’t have to worry about E because we know that D is our choice. I’ll see you soon.