## Video transcript

now let ‘s start to do some more concern problems. And one of these things that you ‘ll find in probability is that you can constantly do a more concern trouble. then now I ‘m going to think about — I ‘m going to take a fair mint, and I ‘m going to flip it three times. And I want to find the probability of at least one principal out of the three flips. So the easiest direction to think about this is how many evenly probably possibilities there are. In the survive television, we saw if we flip a coin 3 times, there ‘s 8 possibilities. For the first flip, there ‘s 2 possibilities. second flip, there’s 2 possibilities. And in the third base flip, there are 2 possibilities. so 2 times 2 times 2 — there are 8 equally likely possibilities if I ‘m flipping a coin 3 times. now how many of those possibilities have at least 1 head ? Well, we drew all the possibilities over here. So we just have to count how many of these have at least 1 head. So that ‘s 1, 2, 3, 4, 5 5, 6, 7. so 7 of these have at least 1 head in them. And this last one does not. so 7 of the 8 have at least 1 head. now you ‘re credibly thinking, OK, Sal. You were able to do it by writing out all of the possibilities. But that would be actually hard if I said at least one head out of 20 flips. This had worked well because I only had 3 flips. Let me make it clear, this is in 3 flips. This would have been a lot harder to do or more time consuming to do if I had 20 flips. Is there some shortcut hera ? Is there some other way to think about it ? And you could n’t equitable do it in some childlike way. You ca n’t just say, oh, the probability of heads times the probability of heads, because if you got heads the first time, then now you do n’t have to get heads anymore. Or you could get heads again — you do n’t have to. So it becomes a fiddling bit more complicated. But there is an easy way to think about it where you could use this methodology right over here. You ‘ll actually see this on a bunch of exams where they make it seem like a unvoiced problem, but if you precisely think about in the correctly way, all of a sudden it becomes simpler. One room to think about it is the probability of at least 1 head in 3 flips is the same thing — this is the same thing — as the probability of not getting all tails, right ? If we got all tails, then we do n’t have at least 1 head. So these two things are equivalent. The probability of getting at least 1 head in 3 flips is the same thing as the probability of not getting all tails in 3 flips. So what ‘s the probability of not getting all tails ? Well, that ‘s going to be 1 minus the probability of getting all tails. The probability of getting all tails, since it ‘s 3 flips, it ‘s the probability of tails, tails, and tails. Because any of the early situations are going to have at least 1 principal in them. And that ‘s all of the other possibilities, and then this is the entirely early leftover possibility. If you add them in concert, you ‘re going to get 1. Let me write it this manner. Let me write it a new coloring material fair therefore you see where this is coming from. The probability of not all tails plus the probability of all tails — well, this is basically exhaustive. This is all of the possible circumstances. indeed your chances of getting either not all tails or all tails — and these are mutually exclusive, so we can add them. The probability of not all tails or, merely to be clear what we ‘re doing, the probability of not all tails or the probability of all tails is going to be equal to one. These are mutually exclusive. You ‘re either going to have not all tails, which means a pass shows up. Or you ‘re going to have all tails. But you ca n’t have both of these things happening. And since they’re mutually exclusive and you ‘re saying the probability of this or this happen, you could add their probabilities. And this is basically all of the possible events. so this is basically, if you combine these, this is the probability of any of the events happening. And that ‘s going to be a 1 or 100 % probability. sol another way to think about is the probability of not all tails is going to be 1 minus the probability of all tails. So that ‘s what we did correct over here. And the probability of all tails is reasonably straightforward. That ‘s the probability of it ‘s going to be 1/2, because you have a 1/2 chance of getting a tails on the beginning flip, times — let me write it hera, so we can have it a little clear. So this is going to be 1 minus the probability of getting all tails. You will have a 1/2 probability of getting tails on the foremost somersault, and then you ‘re going to have to get another tails on the second impudent, and then you’re going to have to get another tails on the third impudent. And then 1/2 times 1/2 times 1/2. This is going to be 1/8. And then 1 minus 1/8 or 8/8 minus 1/8 is going to be adequate to 7/8. So we can apply that to a problem that is harder to do than writing all of the scenarios like we did in the first trouble. Let ‘s say we have 10 flips, the probability of at least one head in 10 flips — well, we use the lapp theme. This is going to be equal to the probability of not all tails in 10 flips. So we ‘re merely saying the probability of not getting all of the flips going to be tail. All of the flips is tails — not all tails in 10 flips. And this is going to be 1 minus the probability of flipping tails 10 times. So it ‘s 1 subtraction 10 tails in a rowing. And thus this is going to be equal to this depart right over here. Let me write this. So this is going to be this one. Let me just rewrite it. This is peer to 1 minus — and this contribution is going to be, well, one dock, another dock. So it ‘s 1/2 times 1/2. And I ‘m going to do this 10 times. Let me write this a little neat. 1/2 — so that ‘s 5, 6, 7, 8, 9, and 10. And indeed we very just have to — the numerator is going to be 1. indeed this is going to be 1. This is going to be equal to 1. Let me do it in that same color of green. This is going to be equal to 1 minus — our numerator, you equitable have 1 times itself 10 times. So that ‘s 1. And then on the denominator, you have 2 times 2 is 4. 4 times 2 is 8, 16, 32, 64, 128, 256, 512, 1,024 — over 1,024. This is the accurate same thing as 1 is 1024 over 1024 minus 1 over 1024, which is equal to 1,023 over 1,024. We have a coarse denominator here. so 1,000 — I ‘m doing that same blue sky — over 1,024. then if you flip a coin 10 times in a quarrel — a fair coin — you ‘re probability of getting at least 1 heads in that 10 flips is pretty high. It ‘s 1,023 over 1,024. And you can get a calculator out to figure that out in terms of a percentage. actually, let me just do that just for fun. so if we have 1,023 divided by 1,024 that gives us — you have a 99.9 % probability that you ‘re going to have at least one heads. So this is if we round. This is equal to 99.9 % opportunity. And I rounded a little bite. It ‘s actually slenderly, even slenderly, higher than that. And this is a pretty brawny joyride or a pretty mighty way to think about it because it would have taken you forever to write all of the scenarios down. In fact, there would have been 1,024 scenarios to write down. so doing this exercise for 10 flips would have taken up all of our time. But when you think about in a slenderly unlike way, when you just say, look the probability of getting at least 1 heads in 10 flips is the lapp thing as the probably of not getting all tails. And that ‘s 1 minus the probability of getting all tails. And this is actually a pretty easy thing to think about.