 Hello and welcome to the session. In this session, first we will discuss about Bernoulli Trials. Each time we toss a coin or roll a dice or perform any other experiment, we call it a trial. The outcome of any trial is independent of the outcome of any other trial. So, trial, the following conditions. The first condition is there should be a finite number of trials This condition is the trials should be to say each trial has exactly a failure. The probability of success at which contains 7 red and 9 black balls. From this bag, we draw 6 balls. Draw the ball drawn is replaced in this case. Now let's see if the trials of drawing the balls are Bernoulli trials or not. As you can see, the number of trials in this case, say drawing a red ball in 7 upon 16, since we have 7 red balls and total number of balls are 16, the of success same for all 6 trials, the distribution. Consider and in this case given ys, so there will be n-x failures of number of successes. The trials may be obtained by the binomial expansion of q plus p whole to the power n which is given as nc0 q to the power n plus nc1 q to the power n minus 1 into p to the power 1 plus nc2 q to the power n minus 2 into p to the power 2 plus and so on up to ncn p to the power n. Thus we have probability distribution of number of successes x is written as given by px values for probabilities for 0 q to the power n. When x equal to 1, then its corresponding probability is nc1 q to the power n minus 1 into p to the power 1, then q to the power n minus 2 into p to the power 2, then when x equal to x, we have probability of x as n minus x into p to the power x, then when x is equal to n, its corresponding probability would be ncn p to the power n. Probability distribution is known as binomial distribution with parameters e we can find the complete probability distribution. The probability of x successes given as into p to the power x where we have x goes from and q is equal to 1 minus p. This function px is called the probability function binomial distribution. So we have a binomial distribution with n Bernoulli trials and probability of success each trial is denoted by b of np. Let us consider an example where we tossed simultaneously. Now in a single throw of a coin, the sample space s would be given by ht if we consider p as the probability of that in this case it would be equal to 1 upon 2 since we get just one head, then q equal to probability would be equal to 1 minus p or equal to 1 minus 1 upon 2 that is 1 upon 2. So we have q equal to 1 upon 2. If we consider the random variable x, then we are supposed to find the probability of getting not more than this would be equal to probability of getting one head or one head or that is equal to probability when x is equal to 0 plus probability when x is equal to 1 plus probability when x is equal to 2 plus probability when x is equal to 3. Now in this formula for the probability function we will substitute the values for nc, q and p to find the probabilities n would be equal to 6 since we have tossed 6 coins. Now to find probability when x is equal to 0 we substitute x equal to 0 in this n equal to 6, q equal to 1 upon 2 and p equal to 1 upon 2 in this formula for probability function. So we get this is equal to 6 c0 1 upon 2 whole to the power 6 plus probability when x is equal to 1 for this we substitute x as 1 this gives us 6 c1 1 upon 2 whole to the power 6 plus probability when x is equal to 2 for this we substitute x equal to 2 which gives us 6 c2 1 upon 2 whole to the power 6 plus probability when x is equal to 3 is given by 6 c3 1 upon 2 whole to the power 6 and this gives us 21 upon 32 a tossed simultaneously probability of getting not more than 3 heads comes out to be equal to 21 upon 32. This completes this session. Hope you have understood the Bernoulli trials and binomial distribution.