 Let us continue with the probability distributions. In the last lecture, we learnt about binomial distribution. Today, we take up another discrete distribution known as Bernoulli distribution Bernoulli distribution. Let us continue with the random experiment of tossing the coin. In that process, we ask a question what is the probability that head will turn up for the first time at the r plus 1th trial. So, we ask a question probability of h turning up for the first time in r plus 1th trial. This means that the first r trials the tails have turned up and therefore, asking what is the probability that at the next trial a head will turn up. If we denote this probability by the notation again p r plus 1 we call this as a Bernoulli distribution. So, I will denote it as p p r it can be worked out as follows. This probability that there has been r cases where tails have turned up successively is equal to r times the probability of individual tail turning up. If q is the probability per trial that a tail will turn up then q to the power r is the probability that successively r times tails have turned up and at the r plus 1th time a head has turned up for which the probability is p and hence we must multiply this quantity by p. So, to speak in words the probability of success for the first time in the r plus 1th trial is q to the power r p for r equal to 0 1 2 etcetera. So, we can understand this better in the convention of considering head as success and tail as failure. So, it is like if one has to appear in some examination and there is no limit you can appear any number of times then you can and if p is the probability of someone passing the examination in each attempt then you can ask the question what is the probability that a person will pass for the first time in the r plus 1th attempt implying that he has failed in the earlier r attempts that probability is once again given by the Bernoulli distribution as given above. It is also called a geometric distribution it is also called geometric distribution because the successive probabilities for r equal to 0 1 2 etcetera follow constant ratios like a geometric progression. There are many examples connected with various disciplines where Bernoulli distribution could be useful. For example, we can have a problem of hazard evaluation. So, supposing it is known that in a well planned city there is a rare probability of flooding occurring once in a way. So, an example of flooding and it is known that flooding occurs if the intensity of the rainfall exceeds some large value I c. So, if the intensity I greater than I c flooding occurs it is also known from experience that the probability of exceedance of rain intensity I beyond I c that probability is given as p let us say per year. Now one can ask a question what is the probability that first time flooding will occur in the r plus 1th year. It has the same formulation that starting from a year where flooding has not occurred the probability of flooding occurring for the first time in the r plus 1th year again once again will be q to the power r p probability of flooding after r years. So, sometimes these probabilities are connected to some often or sometimes called extreme value distributions they are very useful in hazard analysis. So, just an illustration of a very simple example. Let us look at some main properties of Bernoulli distributions. It is a discrete distribution defined on the integer points. So, we have probability of success at r plus 1th trial or attempt called as Bernoulli is q to the power r p where r is 0, 1, 2, etc. And of course, q is 1 minus p as usual. So, which means the probability of success in the first attempt when r equal to 0 is p that is a consistent. This distribution the sum of this distribution that is integral or sum p r plus 1 r varying from 0 to infinity equal to p sigma q to the power r r equal to 0 to infinity e by p by 1 minus q because this is a geometric series whose sum is 1 by 1 minus q and 1 minus q is p. So, it is p by p equal to 1. Hence this probability distribution is normalized. Since there is a finite probability of success at each trial in some trial the person has to succeed for the first time. So, that is why it is 1. We can similarly calculate the mean the average number of attempts to succeed that is defined as sum of r plus 1 p r plus 1 r 0 to infinity and you can show that it is going to be 1 by p. Number of attempts required increases as the probability of success in a each attempt decreases. You can similarly calculate the variance which is r plus 1 square average minus r plus 1 average square and you can it is a bit laborious, but you can work out and show that it is proportional to q by p square or the standard deviation sigma equal to root q by p. If q is fairly close to 1 that is p is close to 0 then we can see that the mean and the standard deviation are of the same order. So, it is also an event where when the success probability is a small the bound the uncertainty bound in the mean value is of the same order as the mean value itself. We can find one more property that is the generating function. The generating function of the binomial distribution defined as z to the power r plus 1 p r plus 1 r equal to 0 to infinity can carry out the sum and it will come to p z divided by 1 minus q. One could have derived the above properties from the generating function also. So, in many problems these Bernoulli distributions will be useful. Actually it is one of the gate ways to understand the extreme value distributions and as and when such situation arises we will discuss it during the course of our study. We next move over to another discrete distribution as known as Poisson distribution. Poisson distribution occurs predominantly in counting statistics and it is a very important method we provides very important methodology for assessing risks for example, due to infections etcetera. It stands on its own footing in many disciplines. So, we dwell at some length on this distribution because of its applications in various contexts. Let us give it an independent example. Consider there is a water reservoir large tank of let us say reasonably clean water, but however there is some bacteria present in this water. Let us say the volume of the tank is large let us say v. Let us say that there are a total of n bacteria in this volume. Let us say somebody takes out a small quantity of water in a glass a volume small v and we now ask a question about the probabilities of the person ingesting I mean if you drinks this water a probability of a person ingesting various numbers of bacteria. This is a very practical question. So, let us now define a quantity nu as the mean number mean concentration of bacteria in water which is since our v is large and n also is expected to be large we call it as n by v. So, if a mean concentration nu exists then the when you collect a small volume of water v the a priori probability that a bacteria could be contained in volume v is basically v by v because a given bacteria originally existed with the probability 1 by v in this entire volume and its probability of being found in small volume v therefore will be v by v and this will happen for each of the n bacterias and an average expectation of the number of bacteria that can be present in the small volume v is a n into v by v. We can also understand this as n by v into v because n by v is the mean concentration. So, if the mean concentration is nu then nu into v is the mean expected number. So, this is expectation or expected number of bacteria in a small volume v. Since v by v is a small and n is large I can write it as n into p as mu. So, now we can see a connection of this with binomial distribution where p is the probability of occurrence of an event here the occurrence of a bacteria. So, it is if that you call a success then the failure is the occurrence non occurrence of a bacteria. So, 1 minus p hence we ask a question probability that the volume v contains r bacteria by a collocation of chances and this should be given by binomial distribution. We know that r can be 0 you are lucky it does not contain any bacteria and one can have a misfortune all the n bacteria also could land up in small volume v because there is an equal probability of bacteria being distributed in the volume v. So, this probability as we saw is p r binomial I show this way is n c r p to the power r 1 minus p or q to the power n minus r this is the stuff we have already done. Now, let us look at the problem a little further supposing my volume is really small and the number of bacteria present in the reservoir are truly large it is a meter cube volume could be millions of bacteria still of course, the concentration could be lower and my volume could be just about 100 m l. So, there is a very little probability v by v is very small say 1 in a million or 1 in 10000. So, in such a situation can I get a limit of the distribution different from the binomial. So, if you have now a constraint in which we say that let the limit n tends to infinity and v tends to 0 the probability n v by v still remaining actually v by v tending to 0 then the limit n v by v equal to mu equal to some fixed number supposing such a situation exists that is the mu equal to n p is fixed in spite of p tending to 0 and tending to infinity. So, that the product is however, fixed let us see what happens to binomial distribution in this limit. So, it is argued later that then the binomial distribution goes over to the Poisson distribution goes over to what is called Poisson distribution that is probability of obtaining R bacteria we call it as Poisson distribution will be equal to that mean number expected mu to the power R divided by R factorial e to the power minus mu unlike the binomial distribution where there were 2 independent parameters that is n and p the Poisson distribution now has only 1 independent parameter mu which is in fact, a limit of product n p where n has tended to infinity and p tends to 0 under this limit. So, it is a 1 parameter distribution how to prove this. So, we now come to the proof of Poisson distribution from binomial distribution under this limiting condition. So, we rewrite the binomial distribution p binomial R in terms of 1 parameter only that will be done by replacing p by mu by n. So, we have this is n factorial by n minus R factorial and R factorial into p to the power R. So, p will be written as mu to the power R by n to the power R and 1 minus p 1 minus mu by n to the power n minus R. We have not done any approximation as of now we have merely replaced p by mu by n. Let us collect terms in a judicious way we can write this probability as p R equal to say n factorial by n minus R factorial and the mu to the power R by R factorial we collect together. So, it will be mu to the power R by R factorial then we will have 1 by n to the power R and the other term we separate as 1 by 1 minus mu by n to the power minus R and 1 minus mu by n to the power n. Let us focus on the last two terms first in the limit n tends to infinity let us see what happens to 1 minus mu by n to the power minus R where R is a number. So, this number tends to 1 that is because R is a fixed number it does not depend on n. So, as n tends to infinity mu by n will tend to 0. So, then 1 to the power any number is 1. So, there is no doubt that this quantity tends to unity. So, one term we have reduced. So, what happens to the other term n tends to infinity 1 minus mu by n to the power n this is not a trivial as you can see as n tends to infinity it is this becomes 1, but 1 to the power infinity is not defined by itself. But we use the binomial theorem which says that 1 plus x by n to the power n as n tends to infinity is e to the power x it is in fact, the definition of exponential. So, accordingly this will tends to e to the power minus mu because x is minus mu here. So, these two terms we have understood we now in the large n limit that is the value it is going to be. So, the remaining terms we write as n factorial by n minus R factorial mu to the power R by R factorial and 1 by n to the power R and e to the power minus mu. Let us now split n factorial we write n factorial as a first R terms that is n into n minus 1 etcetera into n minus R plus 1 and then it will be n minus R factorial only. Now, we note that n factorial by n to the power R can be written for like this with each of these terms divided by n there are exactly those many number of terms in the denominator because there are R terms. So, that in the limit n tends to infinity each of these terms is going to remain 1 because n by n is 1 this will be 1 minus 1 by n n tends to infinity it will be 1. So, this will be 1 minus R plus 1 by n 1 minus R plus R minus 1 by n. So, that also will be 1. So, this will now tend to simply n minus R factorial. There is already one n minus R factorial waiting in the denominator of P R and hence P R now will tend to n minus R factorial divided by n minus R factorial mu to the power R by R factorial e to the power minus mu. So, since this whole term is going to be unity this is mu to the power R by R factorial e to the power minus mu. So, in the limit of n going to infinity and P going to 0 such that mu is fixed the binomial distribution moves over to a Poisson distribution by a process of arriving at the limits. Thank you.