## Video transcript

– [ Voiceover ] Let ‘s say that you know your probability of making a free throw. You know that the probability … The probability, and let’s say of scoring a loose throw because make and miss, they both start with M and that can get confuse, so let ‘s say the probability of scoring … scoring a exempt throw, is equal to, is going to be, say 70 %. If we want to write it as a percentage or we could write it as 0.7 if we write it as a decimal fraction. Let ‘s say the probability of missing a release throw, then and this is equitable going to come directly out of what we merely wrote down, the probability of missing, of missing a complimentary shed, is equitable going to be 100 % minus this. You ‘re either going to make or miss, you ‘re going to score or miss, I do n’t want to use cook in this because they both start with M so this is going to be a 30 % probability, or if we write it as a decimal, 0.3 One subtraction this is 0.7. These are the alone two possibilities, so they have to add up to 100 %, or they have to add up to one. nowadays, let ‘s say that you were going to take six attempts. What we are curious about, is the probability of precisely two scores in six attempts. So let ‘s think about what that is and I encourage you to get inspired at any point in this video you should pause it and you should try to work through what we ‘re asking justly now, so this is what we want to figure out, the probability of precisely two scores in six attempts. then, let ‘s think about the way, let ‘s think about the detail ways of getting two scores in six attempts and think about the probability for any one of those especial ways, and then we can think about how many ways can we get two scores in six attempts ? so, for exercise, you could get you could make the first two release throws, so it could be grade, grudge and then you miss the next four. indeed mark, score, and then it’s miss, miss, miss, and miss. So what ‘s the probability of this exact thing happening ? This exact thing ? Well, you have a 0.7 find of making, of scoring on the first one, then you have a 0.7 chance of scoring on the second one, and then you have a 0.3 casual of missing the adjacent four. sol, the probability of this demand circumstance is going to be what I am writing down. Hopefully you do n’t get the generation symbols confused with the decimals, I ‘m trying to write them a little bit higher. Times 0.3, and what is this going to be equal to ? well, this is going to be equal to … This is going to be 0.7 squared times 0.3 to the one, two, three, fourth — to the one-fourth might. nowadays, is this the lone room to get two scores in six attempts ? No, there ‘s many ways of getting two scores in six attempts. For example, possibly you miss the first one, the first gear undertake and then you make the second try, you score, then you miss the one-third undertake, and let ‘s equitable say you make the fourth attempt, and then you miss the adjacent two. You miss, and you miss. This is another way to get two scores in six attempts, and what ‘s the probability of this happen ? Well, you ‘ll see, it’s going to be precisely this, it ‘s just we ‘re multiplying in a unlike regulate. This is going to be 0.3 times 0.7 you have a 30 % casual of missing the foremost one, a 70 % chance of making the second one, and then times 0.3, a 30 % casual of missing the third, times a 70 percentage chance of making the one-fourth, times a 30 % find for each of the next two misses if you wanted the exact circumstance, this is once again going to be 0.7 and if you just rearrange the order that you ‘re multiplying, this is going to be 0.7 squared times 0.3 to the fourthly power, so for any one of these particular ways to get precisely two scores in six attempts, the probability is going to be this. so, the probability of getting precisely two scores in six attempts, well it ‘s going to be any one of these probabilities times the number of ways you can get two scores in six attempts. well, how … If you have out of six attempts, you ‘re choosing two of them to have scores, how many ways are there ? well, as you can imagine, this is a combinatorics problem, so you could write this as you could write this, let me see how I could … You ‘re going to take six attempts. You could write this as six choose, what we ‘re trying to, you’re picking from six things, six attempts, and you’re picking two of them, or two of them are going to need to be made if you want to meet these circumstances. This is going to tell us the number of different ways you can make two scores in six attempts. Of course, we can write this as kind of a binomial coefficient notation. We can write this is as six, choose two and we can just apply the formula for combinations, and if this looks wholly unfamiliar I encourage you to look up combinations on Khan Academy and then we go into some detail on the argue behind the recipe that makes a draw of common sense. This is going to be equal to six factorial over two factorial times six minus two factorial. Six minus two factorial, I ‘m going to do the factorial in green again, and what ‘s this going to be peer to ? This is going to be peer to six times five times four times three times two, and I ‘ll fair throw in the one there although it does n’t change the value, over two times one. And six minus two is four, so that ‘s going to be four factorial, so this right over here is four factorial so time four times three times two times one. Well, that and that is going to cancel, and the six divided by two is three, so this is 15. There ‘s 15 different ways that you could get two things out of six, I guess, is one way to say it or there ‘s 15 different ways that you could get two things out of six. Another means of thinking about it is there ‘s 15 different ways to make two out of six free throws. now, the probability for each of those is this right over here. The probability of precisely two scores in six attempts, this is where we deserve a little morsel of a drumroll. This is going to be six choose two times 0.7 squared. This is two, you ‘re going to make two and then it ‘s 0.3 to the fourth world power. These will necessarily add up to six. so this right field over here was a three, then this right over here would be a three and then this would be six minus three or three right over here. now, what is this value ? Well, it ‘s going to be equal to we have our 15, three times five so we have this clientele right over here. It ‘s going to be 15 times let ‘s see, in chicken, 0.7 times 0.7 is going to be times 0.49 and let ‘s see, three to the fourthly power would be 81. But, I am multiplying four decimals, each of them have one space to the right field of the decimal fraction item so this is, I ‘m going to have four spaces to the right of the decimal fraction so 0.0081 so there you go, any this total is, and actually I might a well get a calculator out and calculate it, so this is going to be this is going to be … Let me … so, it ‘s 15 times 0.49 times 0.0081 and we get 0.059535 then this is going to be equal to, let me write it down. actually, I wish I had a fiddling act more real number estate right field over here, but I’ll write it in a very bluff color. This is going to be, well, actually, I ‘m kind of out of bold colors. I ‘ll write it in a slenderly less bold color. This is going to be equal to 0.05935 if we wanted the claim count, or we could say this is approximately, if we round to the nearest share this is approximately a six percentage prospect, six percentage probability of getting precisely two scores in the six attempts. I did n’t say two or more, I just said precisely two scores in the six attempts. actually, it ‘s a fairly depleted probability because I have a pretty high absolve throw share. If person has this high of a free throw share it ‘s actually reasonably unlikely that they ‘re merely going to make two scores in the six attempts.

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