 Hello, I am welcome to the session. I am Deepika here. Let's discuss the question which says A card from a pack of 52 cards is lost from the remaining cards of the pack Two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond Now we know that according to Bayes theorem, if even E2 so on E and R events which constitute a partition of sample space S that is even E2 so on E and R pairwise disjoint and Even union E2 union so on union E n is equal to S and A be any event with non-zero probability then probability of EI upon A upon A is equal to probability of EI into probability of A upon EI over sigma probability of Ej into probability of A upon Ej j varying from 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 a given question a card from a pack of 52 cards is lost And from the remaining cards of the pack two cards are drawn and are found to be both diamonds So let even be the event Lost card is diamond to be the event the lost card is not diamond therefore probability of E1 is equal to 13 over 52 and This is equal to 1 over 4 Now in a pack of 52 cards 13 are diamond cards. So 39 are other cards Which are not diamond cards? So probability of E2 is equal to 39 over 52 and this is equal to 3 over 4 a be the event drawing from the remaining cards probability of A upon E1 that is the probability of drawing two diamond cards Given that the lost card is diamond equal to 12 C2 Over 51 C2. This is equal to 12 into 11 over 2 upon 51 into 50 over 2 this is equal to 12 into 11 over 51 into 50 and This is again equal to 132 over 2550 probability of a upon E2 That is probability of drawing two diamond cards Given that the lost card is not diamond. This is equal to 13 C2 over 51 C2. This is equal to 13 into 12 over 51 into 50 This is again equal to 156 over 2550 Now we have to find the probability of the lost card being a diamond so probability lost card being a diamond provided diamond cards are drawn and this is given by probability of even upon a Hence by using Bayes theorem probability of even upon a is equal to probability of even into probability of a upon even over probability of even into probability of a upon even Plus probability of e2 into probability of a upon e2 Now we have probability of even is equal to 1 over 4 probability of e2 is equal to 3 over 4 and probability of a upon even is equal to 132 over 2550 and Probability of a upon e2 is 156 over 2550 So on substituting these values we have probability of even upon a is equal to 1 over 4 into 132 over 2550 upon 1 over 4 into 132 over 2550 plus 3 over 4 into 156 over 2550 this is again equal to 33 over 2550 upon 33 Plus 117 over 2550 and this is equal to 33 over 2550 into 2550 over 150 and this is again equal to 11 over 50 So probability that the lost card being a diamond provided two diamond cards are drawn is 11 over 50 So this is the answer for that question This completes our session. I hope the solution is clear to you. Bye and have a nice day