 probability of an event the probability P of A of an event A from the finite sample space S is given by the ratio between the cardinality of A and the cardinality of S. So it can be written in this way also number of favorable outcomes for the event A divided by number of possible outcomes to find the probability of an event A we sum all the probabilities assigned to the sample points in A this sum is called the probability of A and is denoted by P of A which is like this now from this we can easily say one thing that probability of the whole sample space is always one. So we can write in this way that summation of P of x x belonging to S is equal to 1. So let us consider one example in this context for the experiment of rolling a die the sample space consists of six samples if we suppose that probability of occurrence of any outcome is equal then the probability of getting an odd number is equal to 3 by 6 that is half because there are only six possibilities in the sample space. So sample space contains six sample points so that is why the denominator is 6 and the numerator is 3 because the number of favorable cases will be 1, 3 and 5 because these are the odd numbers available in a die so that is why it will be 3 by 6 which is half. Now let us consider another example it is very very important example let us suppose that a die is loaded so that the numbers through 6 equally likely to appear but that 1 is 3 times as likely as any other to appear. Now to model this situation we should have probability of 2 equal to probability of 3 equal to probability of 4 equal to probability of 5 equal to probability of 6 and another condition is probability of 1 is equal to 3 into probability of 2 since we know that summation p of x x belonging to s is equal to 1 that is why probability of 1 plus probability of 2 plus probability of 3 plus probability of 4 plus probability of 5 plus probability of 6 will be equal to 1. So we can write that 3 into probability of 2 plus probability of 2 plus probability of 2 plus probability of 2 plus probability of 2 that will be equal to 1. So from this we can find the value of probability of 2 which is equal to 1 by 8. So we can find probability of 1 that is 3 by 8. So now if we have to find the probability of an odd number it will be probability of 1 plus probability of 3 plus probability of 5 and that will be 3 by 8 plus 1 by 8 plus 1 by 8 which is 5 by 8. Another very known and important problem we can discuss that is known as birthday problem. So the problem is like this find the probability that among n persons at least 2 people have birthdays on the same date but not necessarily in the same year. Assume that dates equally likely ignore February 29 birthdays. So let denotes the event that at least 2 persons the same birthday. So the event E prime will be the event that 2 persons have same birthday. Now it is easier to compute probability of E prime than probability of E and we can find probability of E from the fact that probability of E plus probability of E prime is equal to 1. Since all months and dates are equally likely and we are ignoring February 29 birthday the size of the sample space will be 365 to the power n sorry n because there are n persons. So that is why the sample space will contain 365 to the power n sample points. The first persons birthday can occur on any one of 365 days. If no 2 persons have the same birthday the second persons birthday can occur on any day except the day of the first persons birthday. Therefore the second persons birthday can occur on any one of remaining 364 days. So first persons birthday can occur in 365 days then second persons birthday can occur on any one of remaining 364 days. Then similarly if we proceed third persons birthday can occur in 363 ways or we can say it can occur on any one of the remaining 363 days. In this way if we proceed we can say that the size of the event that size of the event E prime that no 2 persons have the same birthday no 2 persons have the same birthday. The size of this event will be 365 into 364 into 363 in this way the last one will be 365 minus n plus 1 because there are n persons. Therefore the probability that at least 2 persons have birthdays the same month and date is 1 minus 365 into 363 in sorry 364 into 363 in this way 365 minus n plus 1 divided by 365 to the power n. Now if we consider n equal to 22 so that means if there are 22 persons then this probability will be then the probability is 0.475695. If n is equal to 23 then the probability is 0.507297 thus if n is greater than equal to 23 the probability is greater than half that at least persons birthdays on the same month date. So from this problem this is our conclusion so that is like this if there are more than or equal to 23 persons the probability is greater than half that at least 2 persons have birthdays on the same month and date. Here is another example this is also very interesting example 10 men went to a party and checked their hats when they arrived the hats were randomly returned to them when they departed. We want to know the probability that no man gets his own hat sorry own hat back for the experiment of returning the hats to the men the sample space consists of 10 factorial samples corresponding to the 10 factorial possible permutations of the hats. Let us assume that each permutation occurs with equal probability that is 1 by 10 factorial consequently the probability that no man receives his own hat is equal to 1 by 10 factorial times the number of permutation in which no man receives his own hat let AI event this is an event let AI denotes the set of samples in which the ith man receives his own hat. So the number of favorable cases that at least one person will get his own hat can be written as this cardinality of A1 union A2 union A3 not that union A10. So now how can we find this cardinality so that is the next challenge. So let us consider so we have A1 union A2 A10. So this can be written as summation i equal to 1 to 10 cardinality of AI minus summation i less than j of the cardinality AI intersection A j plus summation i less than j less than k cardinality AI intersection A j intersection A k minus so on. So it will continue in this way the last entry will be minus 1 to the power 10 minus 1 so that will be 9 cardinality of A1 intersection A2 intersection so on intersection A10. So this is coming by inclusion exclusion principle which has already been covered in set theory. So we have like this now what is this summation of the cardinality AI so that means here we have to find the number of ways that one person will get the hat correctly so that means number of ways one person will get his own hat. So let us come to now this slide here summation i equal to 1 to 10 cardinality of AI the number of ways that one person will get his own hat is 10 choose one. So for each of these cases the other arrangements can occur in 9 factorial ways. So that is why the resulting number of ways will be 10 choose 1 into 9 factorial in this way the next one is that that 2 persons will get their own hat and that number is 10 choose 2 and for each of these ways the other arrangements can occur in 8 factorial ways. So the resulting number of ways will be 10 choose 2 into 8 factorial ways. So in this way the last one will be 10 choose 10 0 into 0 factorial. So there 10 people will get their own hat in 10 choose 10 ways. So in this way if we find the cardinality of A1 A2 and so on union A10 then we can find the probability that no man receives his own hat in this way. So it will be 1 by 10 factorial into 10 factorial minus the cardinality of A1 A2 up to union A10. So that cardinality which is nothing but 10 choose 1 into 9 factorial minus 10 choose 2 into 8 factorial up to minus 10 choose 10 into 0 factorial. So the term inside the bracket is coming in this way that number of all possible cases which we have already noted down that is 10 factorial. So 10 factorial minus that cardinality of A1 A2 union A10 will give this term inside this bracket which will give the number of ways that no man receives his own hat and we are dividing it by 10 factorial to get the probability of this event. So it will be 1 minus 1 by 1 factorial plus 2 by 2 factorial minus 3 by 3 factorial and so on up to plus 10 by 10 factorial. So if we calculate this we will get the value 0.36788 now there are three axioms of probability 0 less than equal to probability of A less than equal to 1 for any event A this is the first axiom. Second axiom is probability of a sample space is one which we have already mentioned. Now if A1 A2 A3 A1 A2 A3 and so on is a sequence of mutually exclusive events then probability of A1 union A2 union A3 union like this will be equal to probability of A1 plus probability of A2 plus probability of A3 plus so on. So this is a sum of the probabilities of this events individual probabilities of this events. So the first axiom is obvious second axiom is also obvious we have already mentioned now what about the third one before going to this axiom let us prove one result here. Let E1 and E2 be events then probability of E1 union E2 is equal to probability of E1 plus probability of E2 minus probability of E1 intersection E2. The proof is like this let E1 is equal to x1 x2 to xi E2 is equal to y1 y2 to yj then E1 intersection E2 is considered as z1 z2 to zk. So let us describe this by Venn diagram so it will be like this so this is E1 event this is E2 and this one is E1 intersection E2 so E1 is having the members x1 x2 x3 x4 x5 x6 x7 and E2 is having 6 members y1 y2 y3 y4 y5 and y6 and E1 intersection E2 will be like this it is having 2 members z1 and z2 and z1 is same as x2 which is same as x4 y4 so x2 and y4 are same so those are denoted as z1 and x5 and y5 are same so those these two are denoted by z2 so E1 intersection E2 will contain 2 members z1 and z2 and assume that each set element is listed exactly one time per set then in the list x1 x2 xi y1 y2 to yj z1 z2 to zk so in this list x1 x2 to xi y1 y2 to yj z1 z2 to zk occurs twice so from this it follows that probability of E1 union E2 will be equal to summation probability of x t t equal to 1 to i plus summation probability of yt t equal to 1 to j minus summation probability of zt t equal to 1 to k and this is equivalent to probability of E1 plus probability of E2 minus probability of E1 intersection E2 because we have already seen that z1 z2 to zk occurs if we list x1 x2 to xi y1 y2 to yj so that is why one time we are subtracting it from this list so we will get probability of E1 union E2 so once it is proved then we can consider the third axiom of probability which says that probability of A union B will be probability of A plus probability of B when A and B are mutually exclusive events so this will come from the previous one previous result that is we know that probability of A union B is probability of A plus probability of B minus probability of A intersection B so this result is in general now if A and B are mutually exclusive we know A intersection B will be equal to phi and that is why probability of A union B will be simply probability of A plus probability of B because probability of A intersection B will become 0 in this way if there are three mutually exclusive events A B C then from the inclusion exclusion principle we know that probability of A union B union C is probability of A plus probability of B plus probability of C minus probability of A intersection B minus probability of B intersection C minus probability of A intersection C plus probability of A intersection B B intersection C. So, since A B C are mutually exclusive events all these probabilities will be equal to 0 and that is why we will get A union B union C equal to probability of a plus probability of B plus probability of C. So, let us consider the third axiom here it says that there are mutually exclusive events A 1 A 2 A 3. So, infinite number of mutually exclusive events. So, probability of A 1 union A 2 union A 3 union. So, it will continue. So, this will be the sum of the probabilities individual probabilities of A 1 A 2 A 3 and so on. Because the same reason the probability of the event A 1 intersection A 2 A 1 intersection A 3 will all vanish and that is why we will get only the sum of this probabilities. So, let us discuss now the conditional probability the probability that event A occurs given that event B has already occurred is defined as the conditional probability of event A given the occurrence of event B which is denoted by probability of A given B. So, this for example suppose a die is thrown then if it is asked to find the probability that 4 will appear given that event number has occurred. So, the probability that 4 appeared given that event numbered appeared will be equal to 1 by 3 because event numbers there are only 3 event numbers 2 4 6. So, among this the probability that 4 will occur with probability. So, probability that 4 will occur will be 1 by 3. So, in general let P subscripts P of x i denote the probability associated with example with sample x i given that event B has occurred. Now for x i does not belong to B probability subscripts P of x i will be equal to 0. However for the samples in event B their relative frequencies of occurrence remain the same while the sum of their probabilities should equal to 1 that is summation of probability subscripts P of x i such that this x i belong to B that will be equal to 1. Consequently we need to scale the probability to each of these samples up from probability of x i to probability of x i divided by probability of B. Thus we have probability of x i is equal to 0 when x i does not belong to B and probability of x i divided by probability of B when x i belongs to B. It follows that probability of A given B is equal to summation P B of x i and that is equal to summation x i belonging to A intersection B probability of x i divided by probability of B. Now 1 by probability of B we can take out of this summation. So, we will get 1 by probability of B into summation x i belongs to A intersection B of probability of x i and that is equal to probability of A intersection B divided by probability of B. So, we are getting the definition of conditional probability in this way. So, probability of A given B is probability of A intersection B divided by probability of B. So, let us take one example in this context 3 dice were rolled given that no 2 faces were the same what is the probability that there was an S. Let A denote the event that there was an S and B the event that no 2 faces were the same. Now probability of B will be 6 P 3 that is nothing but 6 into 5 into 4 why is it this the first one. So, what is B actually B the event that no 2 faces were the same. So, the first dice can give 6 outcomes. So, there are 6 possibilities now one outcome has already occurred from the first dice and that can occur in 6 ways. So, once that occurs the second dice will give 5 possible outcomes because the number which has already occurred in the first dice that is discarded from the list. So, that is why it will be 6 into 5 then once the first one and second one have occurred the third dice will give the outcomes in 4 possible ways because the first outcome and second outcome we have to discard from the list. So, there will be 4 outcomes remain. So, that is why we will have total number of outcomes 6 into 5 into 4 which is 6 P 3 and all possible cases what is the number of all possible cases it will be 6 cube. So, 3 dice are there so that is why 6 into 6 into 6 6 cube next we have to find the probability that A intersection B that is both A and B will occur A is the event that there was an S and B the event that no 2 faces were the same. Now if one is already fixed one is an S. So, that is why there will be only 5 possibilities because A and B both will occur. So, that is why that outcome is gone from 2 dices. So, that is why it will be 5 P 2 3 into 5 P 2. So, 3 dice are there. So, for each of this case we will have 5 P 2 possibilities. So, that will be. So, total number of cases will come 3 into 5 P 2 and the denominator is 6 cube because this is the number of all possible cases. So, probability of A intersection B will become 3 into 5 P 2. Thus probability of A given B will become 3 into 5 P 2 divided by 6 P 3 if we calculate this we will get half. So, the conditional probability will be half. So, this is the end of this lecture. Thank you.