 Hello and welcome to the session I am Shashi and I am going to help you with the following question. Question says two groups are competing for the position on board of directors of a corporation. The probabilities that the first and the second group will win are 0.6 and 0.4 respectively. Further if the first group wins the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group. First of all let us understand that if even E2EN are non-empty events which constitute a partition of sample space S and A is any event of non-zero probability then probability of EI then A has occurred is equal to probability of EI multiplied by probability of A upon EI upon summation of probability of EJ multiplied by probability of A upon EJ where J is equal to 1 to N. This theorem is known as Bayes theorem we will use this theorem as our key idea to solve the given question. Let us now start with the solution. Now first of all let us assume let E1 be the event that first group wins. So probability of E1 is equal to 0.6 this is given in the question. Now let us assume that E2 be the event that second group wins. So we can write let E2 be the event that second group wins. Now probability of E2 is equal to 0.4 we are given in the question that probability of winning of second group is equal to 0.4. Now let A be the event of the new product being introduced. Now the probability that the new product is introduced by the group A is equal to 0.7. So we can write probability of A upon E1 is equal to 0.7 and also probability that new product is introduced by second group is given by probability of A upon E2 and it is equal to 0.3 clearly we can see this probability is given in the question. Now we have to find the probability that group 2 introduces the product. So we have to find probability of E2 upon A. Now using Bayes theorem given in key idea we get probability of E2 upon A is equal to probability of E2 multiplied by probability of A upon E2 upon probability of E1 multiplied by probability of A upon E1 plus probability of E2 multiplied by probability of A upon E2. Now we know probability of E1 is equal to 0.6 probability of E2 is equal to 0.4 probability of A upon E1 is equal to 0.7 and probability of A upon E2 is equal to 0.3. Now we will substitute all these values in this expression we get probability of E2 upon A is equal to 0.4 multiplied by 0.3 upon 0.6 multiplied by 0.7 plus 0.4 multiplied by 0.3. Now simplifying further we get 0.12 upon 0.42 plus 0.12 is equal to probability of E2 upon A. Now this further implies probability of E2 upon A is equal to 0.12 upon 0.54 adding these two terms we get 0.54 in the denominator. Now this further implies probability of E2 upon A is equal to 2 upon 9. We will cancel common factor 0.06 from numerator and denominator both and we get required probability is equal to 2 upon 9. So our required answer is probability that the new product introduced was by the second group is equal to 2 upon 9. This completes the session. Hope you understood the solution. Take care and have a nice day.