Let's consider one very good example. And again, as a reminder, these examples on the practice sheet, okay. So here you have an Olympic archer is able to hit the bullseye 80 percent of the time. Assume a shot is independent of the others. If she should six arrows, what's the probability? One that she gets exactly four bullseye. Two, she gets at least four bullseyes and so on. We'll go through all these problems. Now if you notice here you have to analyze the problem to be able to identify it as a binomial kind of experiment. Just in your mind's eye, just picture the Olympic Archer having his or her arrow, okay. And she shoots it. So let's say there's a target. Yeah, and so this eye and arrow, she shoots the arrow. So this is a target, she wants to hit it. Yeah. So right away you see that there are two possible outcomes. If she shoots the arrow, it actually hits the target or she does not. So that can begin to give us a feel that this likely a binomial experiment. Now, let's check independence. If she throws the arrow the first time and she hits the target as the bullseye, is that the reason why she will hit the bullseye the second time, third time, fourth time and so forth? No. So her success, we can see clearly independence here because the, the outcome of the first arrow is now going to determine what the outcome of shooting second arrow would be. So independence, two conditions already satisfied. Now, what about the probability of success? Success here is hitting the bullseye. And so is that probability of success the same throughout? Yes, because this is a this is just one person, one outlet, and she has just one skill. It has skills. And so the probability of hitting the bullseye is 80 percent throughout. And finally, now, want to check does she perform this experiment a fixed number of times. Yes, which our n is six. So all the conditions for a binomial model are satisfied. And so we use the binomial formula to look for all these probabilities. And so when we write it, remember the p is 0.80 and therefore q is 0.20. Remember, q is one minus p, and the number of times she shoots arrows is six, right? Okay, so we need all of this. We're going to introduce a random variable X that counts the number of bullseyes that she hits. So in the first case, number one, let's first of all introduce X number of bullseye. And so we want the probability that she hits exactly four bullseye. So go back here what she gets exactly four bullseyes. So we can write it as out of the six trials, you want exactly four successes and times p probability of success, which is 0.80 raised to the number of successes, four times probability of failure, 20 raised to the number of failures, which is two. Okay? Now we'll use our calculator to break this expression down. Okay? And so right away we go to a binomial PDF, I illustrated that area. And our n is six, our p is 0.80, and the number of successes is four. And that will give us, okay, I already wrote the answer is 0.24623 decimal places 0.246. Okay, let's look at a second one. She gets at least four bullseyes. So at least four means four or greater than four. So at least four bullseyes. So it means she gets four. Or the number of bullseyes is five. Or the normal bullseyes is six. And this is same as the probability that she gets four bullseyes plus the probability that she gets five bullseyes. There's a probability that she gets six bullseyes. Now let you complete this. All you need to do is what is changing here is just the number of successes. So for this first part of the problem we already did in part one. So all you need to do is change the four to five, okay? So this is what I'll do, let me write this completely and I'll let you break that down at home so that you can practice how to use the calculator. Binomial pdf 6.86, okay? And when you break it down, you should have this answer provided 0.901, so 0.901. Okay, let's look at the third part of the problem. Number three, I will split screen here. So you can feed all of that here. In the third part of the problem we're told, she gets at most four bullseyes, at most four. So it cannot exceed four. So in that way, what we're looking at is the probability that X is less than or equal to four. And so she can get 0, [unclear] can get 0 or one, or two, or three. Okay? In this case, we use again the binomial, but this time to be appropriate to use binomial CDF. Okay. [unclear] where it's cumulative. We're not just looking for one. And the inches will be the same. So we'll have six N for N point AT. But now we are looking for probability accumulated probably up to four successes. So I'm going to put four there, okay? So when you want to use or use the binomial CDF, if you have more probabilities to compute, now I did not use it in the previous example. That would've involved some algebra. I didn't just want you to use it right away. But when the number of successes, you're counting the number of successes from 0 to some positive integer, okay, is easier to use the binomial CDF, okay? So this is very straight forward. Rather than looking for the probability that x equals 0, then you add it to the probability that X equals one. And then for x equals two, for x equals three, and for x equals four successes, so you're using one stone and shooting birds. So when you do that, you will have these results. Here at most four that 0.345, 0.345. Okay. Now let's look at the fourth one. She misses the bullseye at least once. And I've drawn your attention to this expression at least once in chapter three, is advisable when you solving probability problems and they ask you to look for the probability of at least one, blah, blah, blah. Then you look for the probability of that outcome not occurring. And then you subtract that from one, that will be the fastest way for you to answer a situation. Now to make this clearer, I will draw the sample space of misses, okay. So if you miss at least once, so you can miss one time. You can miss two times, you can miss three times, you can miss four times, you can miss five times, you can miss six times. Now remember, she shoots six arrows. But then it's possible that if she shoots a sixth arrow, maybe 0 misses. And this is a event that we're more interested in here in this case to, to be able to solve this problem faster. So we're going to avoid solving the problem of looking for the probability of missing one time at a, added to the probability of missing two times, right up to the probability of missing six times. So all we need to do here is look for the probability of no misses, so 0 misses, now subtract that from one. So writing that expression. So probability of missing at least once, same as one minus the probability of 0 misses. And if you have 0 misses, it means you have all successes, right? So she hits a bullseye up to six times. And now from here, you can complete the problem because one minus the probability of six successes is seven minus one minus binomial pdf. So you have eight, sorry six, 0.80 and then six successes. And that probability is give you 0.738, 0.738. Okay, the fifth problem and the six problem, we're told how many bullseye number five do you expect her to get? So we're looking for the expected value here. And in this case, number six, we're looking for the standard deviation. We know that for a binomial model, the expected value for the normal successes is given by n times p. So this is six times 0.80. For number six, we're looking for the standard deviation. So the standard deviation is the square root of npq. So that gives us square root of six times 0.80 times 0.20. And you get the results that we have there. Okay, great. Let's look at one more example. In order to complete our understanding of binomial probability distribution. So assume that 13 percent of people are left-handed. If we select five people at random, find the probability of the, each of the outcome below. So before we look at that, now remember we're talking about people, okay? So you're either left-handed or right-handed, right? So you can see two possible outcomes there. Now if I'm left-handed, that's not a reason why you're left-handed. So being left-handed is independence here. So independent condition is satisfied. Now success here, we're looking for left-handed people. So each time we find a left-handed person, that's success for us. And what's that probability? It's 13 percent. Now remember, we're looking at a large group. So any person you see in the population, the chance of that person being left-handed is 13 percent. Now, how many times are you going to perform the experiment of searching left-handed persons? Well we draw five. Okay, so we have all the ingredients here. So our p probability of success is 0.13. Therefore the probability of failure is 0.87. And a number of times we, number of trials is five. And so the first question, we want to find the probability that in five trials, want to get exactly three lefties, three people who are left-handed. So we can introduce our notation here. So let x be left-handed or number of left-handed persons that we find. So in the first part of the problem, want to find x, when should, find the probability that we find exactly three left-handed persons among the five. So have five choose three. Now probability of success remember is 0.1333 and probability of failure. And this to the two. Again, we shall use our calculators. So we go to binomial pdf. N is five, p is 0.13. And the number of successes we need here is three. Provided an answer here to be 0.0166, so we get 0.0166. In the second part of the problem, we are told, there are at least three lefties. So out of the five persons that we select, we want to get at least three. Okay? And so in this case, we're looking for the probability that X is greater than or equal to three. So x is three, or x is four, x is five. And so now again, you're going to use the binomial function or command, okay? The answer for the first part is already there. So you only need to switch now here is the four and the three becomes four. And in this case also, you will use that function command and three will be replaced by five. And then you add everything. And that will give you, make sure you check that by the answer provided 0.0179, so you get 0.0179. And finally, let's look at the last problem. There are no more than three lefties, no more than three lefties. Again, this is a very good example which we'll use the binomial CDF. So no more than three lefties means x is less than a quarter three. So we cannot choose more than three lefties, we can choose 0. It can be one, two, or three. So let me write that again. So this will be either x is 0, x is one, or x is two, or x is three. Right? Okay. So again, this is a very beautiful problem where we can use the binomial CDF. So we're counting from 0 to some positive integer, and our n is five, our probability of success is 0.13. Moving right up to three successes. And that will give you, again, check that with your calculator, compare your answers. That's 0.9987, so 0.9987. Great. I hope these examples have helped you understand to identify a binomial random variable and also how to calculate probabilities of successes. Here is a self-read example. Okay? So try to do the example and then you compare your answers to the ones I've provided all the solutions here. In our next discussion, we will look at other discrete special discrete probability distributions, namely the Poisson probability distribution and the hyper geometric probability distribution.