The next topic I want to discuss in probability and statistics is probability. And here, first of all, we'll look at the laws of probability and do some examples. And then in the next segment we'll look at Bayes theorem. So, firstly, the laws of probability, definition that probability is the relative likelihood that a particular event will occur. And, as given in the reference handbook, we define P(E) as the probability of some event E occurring. So that P(E) ranges from zero to one, where zero means the event is totally impossible, it will never happen, and one means that the event is inevitable, it will definitely happen. Conversely, we have the probability that the event will not occur is equal to one minus the probability of it occurring. One minus P(E). The next law is the law of total probability, which is the probability that either event A or event B occurs. And that is given by this equation, P(A + B) = P(A) + P(B)- P(A,B), where in this equation as defined here, P of A, is the probability that the event A occurs, P of B, is the probability that B occurs, and P of A,B is the probability that both A and B occur simultaneously. Next we have the law of compound or joint probability, which is the probability that A and B both occur. And that is given by either these expressions, P(A,B), where in this equation P(A) again is the probability that A occurs, and P(A | B) is the conditional probability that B will occur if A has already occurred. So, lets illustrate that with an example. We have a deck of 52 cards, contains three jacks, three queens, and three kings. What is the probability of drawing either a king or a queen from a full deck? So which of these alternatives is it? So in this case we use this equation for this probability that we just had. So in this case, the probability of A occurring, which is the probability of drawing a king. There are 3 kings and 52 cards, so the probability of drawing a king is 3/52, first one here. Similarly, the probability of drawing a queen P(B) is also equal to 3/52 because there are three queens out of 52. And finally the joint probability is zero, because can't draw two cards at the same time here. So therefore, the probability is just the sum of the individual probabilities, 6/52 or 0.115, so the answer is B. Next example, five people gather for a dinner party. The probability that no two of them have the same birthday is which of these alternatives? So, here we'll go through this logically, one step at a time starting with the first person. The probability that the first person is born on any day which is not specified is obviously 100%. It's equal to, excuse me there's 365 days in the year, 365 over 365 is one, or 100%, because we haven't specified what that day is. But now what is the probability that the second person is born on that same day? That probability is only one in 365. Or conversely, the probability that the second person is not born on that day Is much higher, it's 1 minus that, which is equal to 364/365. And we continue in the same way, now that go to the third person. What is the probability that the third person is born on either of those first two days? Well that probability is 2 days out of 365. Or again conversely the probability that the third person was not born on either of those two days is one minus that, or 363/365, etc., and we just carry on to however many people are involved in the group, in this case five. And the joint probability is then the product of all of those individual probabilities. So the probability that no two people have the same birthday is P(1) times the P(2) etc., up to 5 which is equal to (365/365) multiplied, etc. And the answer is .9729, or answer C. In other words, there's a 97% probability that no two people in this group of five have the same birthday. The next example, we have a bag that contains four green socks, two blue socks, and five red socks. Two socks are removed. What is the probability that the first sock that is red and the second one is green? Which of these alternatives? So, this is a joint probability problem and the probability that the first sock is red, which we'll call P(A), is 5 out of 11. In other words, there are 5 red socks, 11 socks total. So the probability that one of them is red is quite high. It's 5 out of 11. The probability that the second sock is green, given that the first one was red, Is P(B | A) is equal to four over ten. In other words, the first one that came out was red, so there's still four green socks left. But now there are only ten socks because we've removed one. So the probability that the second sock is green given that the first one is red, is 4/10. And now, finally, the probability that the first sock is red and the second is green, is just equal to the product of those two. P(A,B) times P(B|A) is equal to 5/11 times 4/10, which is equal to .182. So the answer is D, in other words, there is an 18% probability that the first one was red and the second one was green. And this completes our initial discussion of the laws of probability.
