 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says A bag contains four red and four black balls. Another bag contains two red and six black balls One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag. Let us first understand Bayes theorem. If even E2 so on E n are non-empty events which constitute a partition of sample space s that is even E2 so on E n are pairwise disjoint and even union E2 union so on union E n is equal to s and a is any event of non-zero probability then probability of E i upon A is equal to probability of E i into probability of A upon E i over sigma j varying from 1 to n probability of E j into probability of A upon E j for any i 1 to n. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. According to the question one bag contains four red and four black balls and another bag contains two red and six black balls. One of the two bags is selected at random and a ball is drawn from the bag that is found to be red. We have to find the probability that the ball is drawn from the first bag. Let even be the event of selecting first bag that is bag 1 E2 be the event of selecting second bag that is bag 2 A be the event of drawing a red ball. Then probability of E1 is equal to probability of E2 and this is equal to 1 upon 2 also probability of A upon E1 that is probability of drawing a red ball bag 1 is equal to 4 upon 8 and this is equal to 1 upon 2 probability of A upon E2 that is probability of drawing a red ball second bag and this is equal to 2 upon 8 and this is equal to 1 upon 4. Now we have to find the probability that the ball which was found to be red drawn from the first bag. So the probability drawing a ball first bag being given that it is red is probability of E1 upon A. So by using Bayes theorem we have probability of E1 upon A is equal to probability of E1 into probability of A upon E1 over probability of E1 into probability of A upon E1 plus probability of E2 into probability of A upon E2. Now we know that the probability of E1 and E2 is 1 over 2 and probability of A upon E1 is also 1 upon 2 and probability of A upon E2 is 1 upon 4. So we have probability of E1 upon A is equal to 1 upon 2 into 1 upon 2 over 1 upon 2 into 1 upon 2 plus 1 upon 2 into 1 upon 4 and this is again equal to 1 upon 4 over 1 upon 4 plus 1 upon 8 and this is equal to 1 upon 4 over 2 plus 1 upon 8 and this is equal to 1 upon 4 into 8 upon 3 and this is equal to 2 upon 3. Hence the answer for this question is 2 over 3. So this completes our session. I hope the solution is clear to you. Bye and have a nice day.