Voiceover : Let ‘s say we define the random variable capital X as the act of heads we get after three flips of a bazaar coin. indeed given that definition of a random variable, what we ‘re going to try and do in this video is think about the probability distributions. So what is the probability of the unlike possible outcomes or the different possible values for this random varying. We ‘ll plot them to see how that distribution is spread out amongst those possible outcomes. So let ‘s think about all of the different values that you could get when you flip a fair mint three times. So you could get all heads, heads, heads, heads. You could get heads, heads, tails. You could get heads, tails, heads. You could get heads, tails, tails. You could have tails, heads, heads. You could have tails, heading, tails. You could have tails, tails, heads. And then you could have all tails. So there ‘s eight evenly, when you do the actual experiment there ‘s eight equally probably outcomes hera. But which of them, how would these relate to the value of this random variable ? So lease ‘s think about, what ‘s the probability, there is a site where you have zero heads. So what ‘s the credibly that our random varying adam is equal to zero ? Well, that ‘s this position proper over here where you have zero heads. It ‘s one out of the eight equally likely outcomes. So that is going to be 1/8. What ‘s the probability that our random varying capital X is adequate to one ? Well, let ‘s see. Which of these outcomes gets us precisely one forefront ? We have this one right over here. We have that one right over there. We have this one right over there. And I think that ‘s all of them. then three out of the eight evenly probably outcomes provide us, get us to one heading, which is the same thing as saying that our random varying equals one. sol this has a 3/8 probability. So what ‘s the probability, I think you ‘re getting, possibly getting the hang of it at this point. What ‘s the probability that the random variable adam is going to be equal to two ? Well, for x to be adequate to two, we must, that means we have two heads when we flip the coins three times. So that ‘s this consequence meets this restraint. This consequence would get our random varying to be equal to two. And this result would make our random varying equal to two. And this is three out of the eight evenly likely outcomes. indeed this has a 3/8 probability. And then finally we could say what is the probability that our random variable star x is peer to three ? well, how does our random variable X adequate three ? well we have to get three heads when we flip the coin. So there ‘s only one out of the eight equally likely outcomes that meets that restraint. So it ‘s a 1/8 probability. sol immediately we equitable have to think about how we plot this, to see how this is distributed. So let me draw … so over here on the erect axis this will be the probability. Probability. And it ‘s going to be between zero and one. You ca n’t have a probability larger than one. so just like this. So lashkar-e-taiba ‘s see, if this is one correct over here, and let ‘s see everything here looks like it ‘s in eighths so let ‘s put everything in terms of eighths. So that ‘s half. This is a fourthly. That ‘s a fourth. That ‘s not quite a fourth. This is a fourth right over here. And then we can do it in terms of eighths. So that ‘s a pretty effective approximation. And then over here we can have the outcomes. Outcomes. And thus result, I ‘ll say outcomes for all right let ‘s write this then value for X So X could be zero actually let me do those same colors, X could be zero. ten could be one. x could be two. X could be peer to two. X could be equal to three. X could be equal to three. So these are the possible values for X. And now we ‘re fair going to plot the probability. The probability that X has a value of zero is 1/8. That ‘s, I ‘ll make a short morsel of a stripe veracious over here that goes up to 1/8. so lashkar-e-taiba draw it like this. so goes up to, so this is 1/8 right over here. The probability that X equals one is 3/8. so 2/8, 3/8 gets us correct over let me do that in the purple color So probability of one, that ‘s 3/8. That ‘s right over there. That ‘s 3/8. then let me draw that bar, draw that prevention. And fair like that. The probability that X equals two. The probability that X equals two is besides 3/8. So that ‘s going to be on the same level. Just like that. And then, the probability that X equals three well that ‘s 1/8. So it ‘s going to the lapp altitude as this thing over here. I ‘m using the wrong discolor. So it ‘s going to look like this. It ‘s going to look like this. And actually let me just write this a little spot neat. I can write that three. Cut and paste. motion that three a fiddling closer in sol that it looks a little bit clean. And I can actually move that two in actually adenine well. So cut and spread. So I can move that two. And there you have it ! We have made a probability distribution for the random variable x. And the random variable x can only take on these discrete values. It ca n’t take on the value one-half or the value private detective or anything like that. so this, what we ‘ve just done here is constructed a discrete probability distribution. Let me write that down. So this is a discrete, it only, the random variable only takes on discrete values. It ca n’t take on any values in between these things. so discrete probability. Probability distribution. distribution for our random variable ten.