 Hi and how are you all today? The question says three urns A, B and C contain six red and white, two red and white and one red and white balls respectively. An urn is chosen at random and a ball is drawn. If the ball drawn is found to be red, find the probability that the ball was drawn from urn A. So here let E1 be the event of selecting urn A. Let E2 be the event of selecting urn B. And let E3 be the event of selecting urn C. Also let A be the event that the ball drawn is red. Right? Also we are given that probability of E1 is equal to probability of E2 is equal to probability of the third event will be equal to they are having equal chances. So it will be 1 by 3. Probability of drawing a red ball from urn A that is P, A upon E1 is equal to there are six red balls in urn A. So it is 6 by in total there are 10 balls. Similarly probability of selecting a red ball from urn 2 will be equal to 2 by 8 and the probability of selecting a red ball from urn 3 is equal to 1 upon 6 as there are 1 red ball in urn 3 sorry urn C which is denoted by E3 and there are in total 6 balls in this urn. Now by Bayes theorem we can say that the probability of selecting first of all urn A and then a red ball from there will be equal to probability of selecting urn A into probability of selecting red ball from this urn divided by probability of urn 1 sorry A into probability of A upon E1 plus probability of that is the total E2 into probability of A upon E2 plus probability of E3 into probability of A upon E3 right. So now substitute all the values and on doing so we have 1 upon 3 that is probability of choosing urn A into 6 upon 10 this value we have found out above upon 1 upon 3 into 6 by 10 plus 1 upon 3 into probability of A upon E2 is 2 upon 8 plus 1 upon 3 into 1 by 6. So we have it as 36 upon 61 after solving it and this is the required answer to the question which is given to us. So hope you understood it well and enjoyed it too have a nice day.