 Hello and welcome to the session. I am Deepika and I am going to help you to solve the following question. The question says, An arm contains seven red and four blue balls. Two balls are drawn at random with replacement. Find the probability of getting two red balls, three blue balls, one red and one blue ball. So let's start the solution. Now we are given an arm contains seven red and four blue balls. So we have the total number of balls is equal to seven plus four, which is again equal to eleven. So we have the probability of a red ball is equal to seven over eleven and probability of a blue ball is equal to four over eleven. Again according to the question, two balls are drawn at random with replacement. Now in part A we have to find the probability of getting two red balls. Now since the balls are drawn with replacement and probability of a red ball is equal to seven over eleven, therefore the probability of drawing two red balls is equal to seven over eleven into seven over eleven, which is further equal to forty nine over one twenty one. Now in part B we have to find the probability of getting three blue balls, since probability of a blue ball is equal to four over eleven, therefore the probability of three blue balls is equal to four over eleven into four over eleven into four over eleven. And this is further equal to sixty four over one thousand three hundred thirty one. Now we have to find the probability of one red and one blue ball. So the probability of one red ball and one blue ball is equal to seven over eleven into four over eleven plus four over eleven into seven over eleven. Because if the first ball drawn is red and the second ball drawn is blue, then its probability is seven by eleven into four by eleven. But if the first ball drawn is blue and the second ball drawn is red, then its probability is four over eleven into seven over eleven. So the probability of one red and one blue ball is equal to seven over eleven into four over eleven plus four over eleven into seven over eleven. And this is further equal to twenty eight plus twenty eight over one twenty one. And this is again equal to fifty six over one twenty one. Hence the probability of two red balls is forty nine over one twenty one. And probability of three blue balls is equal to sixty four over one thousand three hundred thirty one. And probability of one red and one blue ball is fifty six over one twenty one. So this is the answer for the above question. This completes our session. I hope the solution is clear to you. Bye and have a nice day.