 Hello and welcome to the session. In this session we discussed the following question which says three bags contain balls as shown in the table below. So this table shows the bag number, the number of white balls, number of black balls and number of red balls in the respective bags. A bag is chosen at random and two balls are drawn from it. They happen to be white and red. What is the probability that they came from third bag? Before we move on to the solution, let's recall the Bayes theorem, according to which we have F, even, E2 and so on up to En are n non-empty events which constitute the partition of sample space S that is we have even E2 and so on up to En are device disjoint and we have even union E2 union and so on union En is equal to S and we have A is any event of non-zero probability. Then probability of EI given that event A has already occurred is equal to probability of EI into probability of A such that EI has already occurred. This whole upon summation probability of EJ into probability of A such that EJ has already occurred and summation J goes from 1 to N for any I is equal to 1, 2, 3 and so on up to N. This is the key idea that we use for this question. Let's move on to the solution now. First of all we take let A1 be the event of choosing bag 1, A2 be the event of choosing bag 2 and A3 be the event of choosing bag 3. We assume that E be the event of choosing 2 balls, 1 white and 1 red from a bag. Now since we have 3 bags with us so now probability of choosing bag 1 that is probability of A1 would be same as the probability of choosing bag 2 that is probability of A2 and this would be same as the probability of choosing bag 3 that is probability of A3 and this would be equal to 1 upon 3. Now the probability of choosing 2 balls that is 1 white and 1 red ball from the bag 1 that is probability of E given that A1 has already occurred is equal to choosing 1 white ball from the bag 1 Now from the table you can see that we have 1 white ball in the bag 1 so it would be 1C1 into choosing 1 red ball from the bag 1. Now from the table we have 3 red balls so out of 3 we would choose 1 red ball so it would be 3C1 upon choosing 2 balls from the total balls in the bag 1. Now from the table you can see that the total balls in the bag 1 are 6 so in the denominator here we would have 6C2 so this would be equal to 1 into 3 upon 15 that is 3 upon 15 and this is equal to 1 upon 5 so probability of choosing 2 balls 1 white and 1 red ball from the bag 1 is equal to 1 upon 5 Now in the same way let's find the probability of choosing 2 balls 1 red and 1 white from the bag 2 that is probability of E given that A2 has already occurred would be equal to now since there are 2 white balls in the bag 2 so this would be equal to 2C1 into now there is 1 red ball in the bag 2 so here we would have 1C1 upon 2 balls from the bag 2 now there are total of 4 balls in the bag 2 so here we would have 4C2 and this is equal to 2 into 1 upon 6 now 2 3 times is 6 so this is equal to 1 upon 3 that is probability of choosing 2 balls 1 white and 1 red from the bag 2 is equal to 1 upon 3 now probability of choosing 2 balls 1 white and 1 red from the bag 3 is equal to choosing 1 white ball from the bag 3 now since there are 4 white balls in the bag 3 so here we would have 4C1 into now there are 2 red balls in the bag 3 so out of 2 we would have 1 red balls it would be 2C1 upon total number of balls in bag 3 that is 9 and we choose 2 balls from these 9 balls it would be 9C2 and so this is equal to 4 into 2 upon 36 now this 4 9 times is 36 so this is equal to 2 upon 9 so now we have probability of event E given that A3 has already occurred or you can say probability of drawing 2 balls 1 white and 1 red from the bag 3 is equal to 2 upon 9 so we need to find probability of A3 given that E has already occurred that is probability of choosing bag 3 for drawing the balls given that the balls drawn are 1 white and 1 red is equal to probability of A3 multiplied by probability of E given that A3 has already occurred that is probability of choosing 2 balls 1 white and 1 red from the third bag this whole upon submission probability of AI multiplied by probability of E given that AI has already occurred that is probability of choosing 2 balls 1 white and 1 red from the 3 bags A1 A2 or A3 and I goes from 1 to 3 now probability of A3 is 1 upon 3 so we have this is equal to 1 upon 3 multiplied by 2 upon 9 this whole upon probability of A1 that is we put I equal to 1 and that is 1 upon 3 into probability of E given that A1 has already occurred that is 1 upon 5 that is probability of choosing the 2 balls 1 white and 1 red from the bag 1 is 1 upon 5 plus probability of A2 1 upon 3 into probability of choosing 2 balls 1 white and 1 red from the bag 2 and that is equal to 1 upon 3 plus probability of A3 that is 1 upon 3 into probability of choosing 2 balls 1 white and 1 red from bag A3 and that is 2 upon 9 so this is equal to 1 upon 3 into 2 upon 9 and this whole upon 1 upon 3 common from the denominator and inside the bracket we have 1 upon 5 plus 1 upon 3 plus 2 upon 9 now this 1 upon 3 1 upon 3 cancels and so this is equal to 2 upon 9 upon 3 we take the LC1 as 45 then here in the numerator we have 9 plus 15 plus 10 and so this is equal to 2 upon 9 upon 34 upon 45 so we get this is further equal to 2 upon 9 into 45 upon 34 by lying 5s are 45 and 2 17 times is 34 so this is equal to 5 upon 17 that we get that we get that the probability that the balls are drawn from bag 3 that is bag A3 given that the balls drawn are 1 white and 1 red is equal to 5 upon 17 so this is our final answer this completes the session hope you have understood the solution of this question.