Hi and welcome back. In this video, we're going to continue our learning goals for Module 1. In particular, we're going to understand permutations and combinations and be able to calculate probabilities when each symbol event is equally likely. Recall that the goal of probability is to assign some number that we're calling P of A, and we want to give a precise meaning to the chance that A will occur. If a sample space S has N single events, and if each of these events is equally likely to occur, then we only need to count the number of events to find the probability. For example, if S is E1, E2 up through EN, those are all the events in the sample space and if each one of those events has the same probability, 1 over N, then if A is some subset of S, we can calculate the probability of A by counting the number of simple events in A and dividing by N. Let's do an example. Suppose our experiment is to roll a single six-sided dice twice. I'm going to indicate the sample space by all the pairs i, j, where i is the role on the first day, and j is the role on the 2nd. Since each one has six possible outcomes, my sample space is going to have 36 events, I'm going to assume that the dice are fair so that each one of those 36 outcomes is equally likely. For example, if I let A be the event of rolling a 1 on the first roll, then the probability of A is the probability of 1 1, 1 2, 1 3, 1 4, 1 5, and 1 6. That event has probability one-sixth. If B is the event that the sum of the two roles is eight, that would be the probability of a 2 and a 6, a 3 and a 5, 4 4, 5 3, and 6 2. There are five events there. The probability is 5 over 36. Finally, as another example, let C be the event that the value of the second role is two more than the first row. That would be the probability of rolling a 1 on the first, and then a 3, a 2 and a 4, a 3 and a 5, or a 4 and a 6, there are four possibilities, so the probability is 4 out of 36 or one-ninth. Let's take a slightly different or a slightly more complicated example, suppose I have an organization that has 60 members, I'm going to pick one person at random to be the president, another person to be the vice president, and a third person to be the treasurer. How many ways can this be done? Let's think about that. The first person I'm picking out of all 60 people. I have 60 choices for the first person. Then I only have 59 people left. I have 59 choices for the second person who's going to be the vice president and then I have 58 choices for the third person who's going to be the treasurer. That would be the number of permutations of size three. Any ordered sequence of k objects taken from a set of n objects is called a permutation of size k. In a permutation order matters. It matters who I'm picking as the first person to be the president versus picking that same person to be the vice president. Order is important. I want a shorthand notation for writing this and my shorthand notation is going to be P sub 3, 60 or in general P sub k, n. I also want to notice that I could rewrite these using factorials. This is the same as 60 factorial over 57 factorial. You'll see in the next slide why this is going to be helpful for us. If I really wanted to know the number, I could throw it into a calculator and I would get 205,320, although that's less important. To remind you about factorials and factorial is n times n minus 1 times n minus 2, all the way down to 1 and I do that for any positive integer n. By definition and for convenience we take 0 factorial to be equal to 1. Now, suppose instead I want to take those same 60 people, but instead of choosing a president, vice president and treasurer, I just want to choose a team of three people and then order is not important. How many combinations, how many ways are there to do this? Well, let's go back to our permutations and think about the sample space, sample space of permutations. I'm going to put a number on all 60 people, so I have a convenient way of putting them into the sample space. I could choose one, two, three. Persons one, two and three, the first person being the president, then the vice president, then the treasurer. Or I could have taken those same three people and I could have made the first person, the president, but I could have made the third person, the vice president and the second the treasurer. Similarly 213, 231, 312, and 321. Those same three people could have been rearranged as president, vice president and treasurer, or I could have taken the first two people and then the fourth person and I could have arranged those six different ways like that, or I could have taken the first two people and the fifth person and so on, this would have given me the sample space of all 205,320 permutations, but I don't want all the permutations, what I want to do is I want to group these into combinations. I'm going to take these six people and I'm going to call that one combination and then I'm going to take these six permutations and group that into one combination and so I'm going to do that all the way down. Instead of having a sample space of permutations, I'm going to have a sample space of combinations and I will denote that. Combinations, let's put a C for combination's and a P here for permutations just to keep it straight in this particular slide. I would have, 123, 124, 125 and so on for my combination's. In general, given N distinct objects, any unordered subset of size K of the objects is called a combination, so our notation is going to be C for combination's. In this particular case, I want you to notice that the cardinality of the permutations, which we saw on the previous slide was 60 factorial, over 57 factorial here for the combinations, the cardinality is going to be the same, 60 factorial over 57 factorial but we're condensing all six here into one. I want to divide by three factorial, which is the same as six. Let's rewrite that so the cardinality for combination's is my 60 factorial, 57 factorial and my three factorial. This comes up so frequently we want a specific notation for it, and the notation we're going to use is 60, choose three. The number of combinations of three people chosen from a group of 60 distinct objects is going to be 60 choose three. In general, our notation is and choose k n factorial divided by K factorial times N minus K factorial. I will observe, just note, 60 choose three is exactly the same as 60, choose 57. Think about it this way, if I do 60, choose three, I'm choosing three people to be in my group and then there's 57 people that are left and you could think of that as a different group, so then 60 choose 57 is exactly the same number. Let's do another example. Same group of 60 people, but now I'm going to add in one more piece of information. Thirty-five of those people are women, and 25 are men. Now, instead of three people, I want to select a committee of 11 people. How many ways can such a committee be formed? The number of committees of 11 would be 60, choose 11. That number is going to be quite large, so I'm not going to multiply it out. I'm just going to leave it as 60, choose 11, realizing that this is 60 factorial over 11 factorial times 60 minus 11 factorial is also, by the way. The cardinality of our sample space. Now, what is the probability that a randomly selected committee will contain at least five men and at least five women? We're going to assume each committee of 11 people is equally likely. With that thought in mind, the probability of at least five men and at least five women on the committee is going to equal the probability of five men and six women plus the probability of six men and five women. To figure that out, think about our 35 women in one group, and our 25 men in another. We're going to have 25 men, and I want to choose five of them. I have 35 women and I want to choose six of them. The sample space has 60, choose 11. Then I'm adding to that, the probability of 25 men choosing six of them and 35 women, and choosing five over the same 60 choose 11. This will give the probability of finding that men and women on your committee. Let's do one more example, suppose your city has bought 20 buses, and shortly after being put into service, some of them develop cracks in the frame, the buses are inspected and eight have visible cracks. You want to select a sample of five for more thorough inspection. This problem, even though it's buses and cracks in the frame, this is exactly the same as the previous problem where we had men and women on our committee. Here we have buses with cracks and buses without visible cracks, and I want to choose five for a thorough inspection. My sample size is going to be 20 choose five. I'm assuming that each bus is distinguishable, it's distinct from every other bus. If five buses are chosen at random, find the probability that exactly four have cracks. The probability of four with cracks is going to be, I have 12 buses without cracks, so I have to choose one of those, and I have eight buses with cracks and I have to choose four of those divided by 20, choose five. What about if I want to ask for the probability that at least four have cracks? Well, that would be the probability of exactly four with cracks and the probability that five have cracks. The probability of at least four with cracks is going to be the probability of exactly four, so that would be 12, choose one, eight, choose four, divided by 20, choose five, plus the probability 12, choose zero, eight choose five, divided by 20, choose five. Hopefully, this helps you understand the difference between permutations, and combinations and how to calculate probabilities in those situations. In the next video, we're going to extend this to conditional probabilities. We'll see you then.
