I ‘m going to start with a fair coin, and I ‘m going to flip it four times. And the first base question I want to ask is, what is the probability that I get precisely one promontory, or heads ? This is one of those confuse things, when you ‘re talking about what side of the coin. I know I ‘ve been not doing this systematically. I ‘m tempt to say, if you ‘re saying one, it feels like you should do the singular, which would be head. But I ‘ve read up a fiddling bit of it on the internet, and it seems like when you ‘re talking about coins, you very should say one heads, which is a short bit, it seems a little snatch unmanageable for me. But I ‘ll try to go with that. So what is the probability of getting precisely one heads ? And I put that in quotes to say, you know, very, we ‘re just to talking about one steer there. But it ‘s called heads when you ‘re dealing with coins. Anyway, I think you get what I ‘m talking about. And to think about this, let ‘s think about how many different potential ways we can get four flips of a coin. So we ‘re going to have one flip, then another flip, then another interchange, then another flip. And this first flick has two possibilities. It could be heads or tails. The second flip has two possibilities. It could be heads or tails. The third base flip has two possibilities. It could be heads or tails. And the one-fourth flip has two possibilities. It could be heads or tails. thus you have 2 times 2 times 2 times 2, which is equal to 16 possibilities. 16 potential outcomes when you flip a coin four times. And any one of the possible outcomes would be 1 of 16. So if I wanted to say, so if I were to merely say the probability, and I ‘m just going to not talk about this one pass, if I barely take a, precisely possibly this thing that has three heads veracious here. This exact sequence of events. This is the first flip, second flip, third gear flip, one-fourth flip. Getting precisely this, this is precisely one out of a possible of 16 events. now with that out of the way, let ‘s think about how many possibilities, how many of those 16 possibilities, involve getting precisely one heads ? well, we could list them. You could get your heads. So this is equal to the probability of getting the heads in the first pass, plus the probability of getting the heads in the second somersault, plus the probability of getting the heads in the third base flip. Remember, precisely one heads. We ‘re not saying at least one, precisely one heads. So the probability in the third gear flip, and then, or the possibility that you get heads in the fourthly flip. Tails, heads, and tails. And we know already what the probability of each of these things are. There are 16 possible events, and each of these are one of those 16 possible events. therefore this is going to be 1 over 16, 1 over 16, 1 over 16, and 1 over 16. And sol we ‘re truly saying the probability of getting precisely one heads is the same thing as the probability of getting heads in the first flip, or the probability of getting heads — or I should say the probability of getting heads in the first flip, or heads in the second flip, or heads in the third flip, or heads in the fourthly flip. And we can add the probabilities of these different things, because they are mutually exclusive. Any two of these things can not happen at the like time. You have to pick one of these scenarios. And so we can add the probabilities. 1/16 plus 1/16 plus 1/16 plus 1/16. serve I say that four times ? Well, assume that I did. And so you would get 4/16, which is peer to 1/4. Fair adequate. now let ‘s ask a slenderly more matter to question. Let ‘s ask ourselves the probability of getting precisely two heads. And there ‘s a couple of ways we can think about it. One is barely in the traditional way. [ ? We ? ] know the number of possibilities and of those equally likely possibilities. And we can entirely use this methodology because it ‘s a fair coin. sol, how many of the sum possibilities have two heads of the total of evenly likely possibilities ? So we know there are 16 equally likely possibilities. How many of those have two heads ? indeed I ‘ve actually, ahead of time so we save time, I ‘ve drawn all of the 16 equally probably possibilities. And how many of these involve two heads ? well, let ‘s see. This one over here has two heads, this one over here has two heads, this one over here has two heads. Let ‘s see, this one over here has two heads, and this one complete hera has two heads. And then this one over here has two heads, and I believe we are done after that. indeed if we count them, one, two, three, four, five, six of the possibilities have precisely two heads. so six of the 16 evenly likely possibilities have two heads. So we have a — what is this — a 3/8 chance of getting precisely two heads. now that ‘s kind of what we ‘ve been doing in the past, but what I want to do is think about a way so we would n’t have to write out all the possibilities. And the reason why that ‘s useful is, we ‘re only dealing with four flips now. But if we were dealing with 10 flips, there ‘s no room that we could write out all the possibilities like this. So we truly want a different room of thinking about it. And the different way of thinking about it is, if we’re saying precisely two heads, you can imagine we’re having the four flips. Flip one, flip two, flip three, flip four. So these are the flips, or you could say, the result of the flips. And if you ‘re going to have precisely two heads, you could say, well, search, I ‘m going to have one head in one of these positions, and then one head in the other position. So how many, if I’m picking the first, therefore I have kind of a heads one, and I have a heads two. And I do n’t want you to think that these are somehow the heads in the first gear flip or the heads in the moment somersault. What I ‘m saying is we need two heads. We need a sum of two heads in all of our flips. And I ‘m just giving one of the heads a name, and I ‘m giving the other head a name. And what we ‘re going to see in a few seconds is that we actually do n’t want to double count, we do n’t want to count the site, we do n’t want to double count this situation — heads one, heads two, tails, tails. And heads two, heads one, tails, tails. For our purposes, these are the exact same outcomes. So we do n’t want to double count that. And we ‘re going to have to account for that. But if we precisely think about it broadly, how many different spots, how many different flips can that first lead show up in ? Well, there ‘s four different flips that that beginning head could show up in. So there ‘s four possibilities, four flips, or four places that it could show up in. Well, if that beginning mind takes up one of these four places, let ‘s barely say that first forefront shows up on the third base flip, then how many unlike places can that second principal show up in ? Well, if that first head is in one of the four places, then that moment head can merely be in three different places. So that moment head can lone be — I ‘m picking a dainty color here — can merely be in three different places. And so, you know, it could be in any one of these. It could possibly be correct over there. Any one of those three places. And then, when you think about it in terms of the first, and I do n’t want to say the first head, head one. actually, let me call it this manner. Let me call it head A and principal B. That way you wo n’t think that I’m talking about the first flip or the second flip. So this is head A, and this good over there is question B. therefore if you had a especial, I mean, these heads are identical. These outcomes aren’t different, but the way we talk about it right field now, it looks like there ‘s four places that we could get this forefront in, and there ‘s three places where we could get this head in. And therefore if you were to multiply all of the different ways that you could get, all of the different scenarios where this is in four different places, and then this is in one of the three left over places, you get 12 different scenarios. But there would only be 12 different scenarios if you viewed this as being different than this. And let me rewrite it with our new — So this is head A, this is head B, this is fountainhead B, this is head A. There would only be 12 different scenarios if you viewed these two things as basically different. But we do n’t. We ‘re actually double count. Because we can constantly swap these two heads and have the demand same consequence. So what you want to do is actually divide it by two. So you want to divide it by all of the unlike ways that you can swap two different things. If we had three heads here, you would think about all of the different ways you could swap three unlike things. If you had four heads here, it would be all the different ways you could swap four different things. So there ‘s 12 different scenarios if you couldn’t swap them, but you want to divide it by all of the different ways that you can swap two things. so 12 divided by 2 is equal to 6. Six different scenarios, basically different scenarios, considering that you can swap them. If you assume that head A and question B can be interchangeable. But it ‘s a wholly identical result for us, because they’re truly just heads. So there ‘s six different scenarios, and we know that there ‘s a entire of 16 evenly likely scenarios. So we could say that the probability of getting precisely two heads is 6 times, six scenarios and — Or there’s a couple of ways. You could say there are six scenarios that give us two heads, of a potential 16. Or you could say there are six possible scenarios, and the probability of each of those scenarios is 1/16. But either way, you’ll get the lapp answer.
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