 Hello and welcome to the session. In this session, we discuss the following question which says a die thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die. Before we move on to the solution, let's recall the binomial distribution according to which we have probability of x equal to x is equal to mcx into p to the power x into q to the power m minus x where we have x is equal to 0, 1, 2 and so on up to n. Here we have m is equal to the number of trials in an experiment. Then this capital X is the random variable. Small x is the number of successes. Then the small p is the probability of getting a success in a single trial and q is the probability of getting a failure in a single trial. And we have p plus q is equal to 1. This is the key idea that we use in this question. Let's move on to the solution now. Now in the question it says that we need to find the probability of obtaining the third six in the sixth throw of the die. So we have the probability of obtaining in the sixth throw of the die would be equal to the probability of obtaining in first five throws and one six in. So this means this would be equal to the probability of obtaining sixes. In first five throws multiplied by the probability of obtaining one six. Now the probability of obtaining one six in the sixth throw would be equal to one upon six. Let this be equal to small p. So small p is the probability of getting a success which is probability of obtaining one six in the sixth throw and this is equal to one upon six. Now we know that p plus q is equal to one where this q is the probability of getting a failure and from here we get q is equal to one minus p. Putting the values at p we get one minus one upon six. So we get q is equal to five upon six. So we have p equal to one upon six and q equal to five upon six. Now let's find out the probability of obtaining sixes would be equal to using the binomial distribution in the key idea we get this would be equal to mcx into p to the power x into q to the power n minus x. Now here we would take n that is the number of trials would be five since we are doing five throws. x is the number of successes and in this case it would be two since we would be obtaining two sixes and that is a success. So x would be equal to two. The small p is the probability of obtaining one six in the sixth throw which is equal to one upon six. That is small p is equal to one upon six and q is equal to five upon six. Putting the respective values in this we get the probability of obtaining sixes in first five throws would be equal to five c2 into one upon six to the power two into five upon six to the power n minus x which would be five minus two that is three and so this is equal to factorial five upon factorial two into factorial three into one upon six into one upon six into five upon six into five upon six into five upon six. So further we get this is equal to ten that is solving this we get ten multiplied by 125 upon seven seven seven six further this is equal to one two five zero upon seven seven seven six. So this is the probability of obtaining two sixes in the first five throws that is one two five zero upon seven seven seven six. Now probability of obtaining third six in the sixth row of the die is equal to probability of obtaining two sixes in first five throws multiplied by probability of obtaining one six in the sixth throw. So now we have the probability of obtaining third six in the throw is equal to probability of obtaining one six in the 6th throw which is equal to P that is 1 upon 6 multiplied by probability of obtaining two sixes in first five throws that is 1, 2, 5, 0 upon 7, 7, 7, 6 so this is equal to 1, 2, 5, 0 upon so 6, 6, 5, 6 now 2, 6, 25 times is 1, 2, 5, 0 and 2, 2, 3, 3, 2, 8 times is 4, 6, 6, 5, 6 and so this is equal to 6, 25 upon 2, 3, 3, 2, 8 so finally we get the probability of obtaining third six in the sixth throw is equal to 6, 25 upon 2, 3, 3, 2, 8 so this is our final answer this completes the session hope you have understood the solution of this question.