 Hello and welcome to the session. In this session, we are going to discuss the following question which says that 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. A question says that a card is lost from a pack of 52 cards. And from the remaining cards of the pack, two cards are drawn and are found to be both diamonds. And we need to find the probability of the lost card being a diamond card. As we have already stated, if an event A occurs, given that an event B has already occurred, then the conditional probability of A when B has already occurred is given by probability of event A when event B has already occurred is equal to probability of A intersection B upon probability of B where probability of B is non-zero. So here, we will be using Bayes' theorem which states that if E1, E2 and so on up to Em is at the top and mutually exclusive and collectively exhausted events. And if A be any event such that probability of event A is not equal to zero, then probability of event EI when event A has already occurred is equal to probability of event EI into probability of event A when event EI has already occurred upon summation of probability of event Ej into probability of event A when event Ej has already occurred where J goes from 1 to M also a set of events E1, E2 and so on up to Em I said to be mutually exclusive if EI intersection Ej is empty for all I on J where I is not equal to J and a set of events E1, E2 and so on up to Em I said to be collectively exhausted if E1, E2 and so on up to Em is equal to the sample space S with this the idea we are going to proceed with the solution As we are given in the question that a card from a pack of 52 cards is lost, let E1 be the event that the lost card is diamond E2 be the event that the lost card is not diamond and we know that in a pack of 52 cards there are 13 diamonds, 13 space, 13 hearts and 13 flecks so out of the 52 cards there are 13 diamond cards, 39 are other cards so probability of event E1 that the lost card is diamond is given by 13 upon 52 as out of 52 cards 39 diamond cards and is given by 1 by 4 also probability of event E2 that the lost card is not diamond is given by 39 upon 52 as out of 52 cards there are 39 other cards which are not diamond cards and it is given by 3 by 4 now let A be an event of drawing 2 diamond cards from the remaining cards now as event E1 is that the lost card is diamond therefore probability of event A when event has already entered is equal to probability of drawing 2 diamond cards given that the lost card is diamond is equal to 12 C2 upon 51 C2 as out of 14 diamond cards if the lost card is diamond then the 2 cards are drawn from the remaining 12 diamond cards now as we know that NCR is given by N factorial upon I factorial into N minus I factorial therefore by using this formula in 12 C2 where N is equal to 12 and R is equal to 2 we get 12 factorial upon 2 factorial into 12 minus 2 factorial that is 10 factorial whole upon similarly we can use the same formula in 51 C2 where N is equal to 51 and R is equal to 2 and we get 51 factorial upon 2 factorial into 51 minus 2 factorial that is 39 factorial which is equal to 12 factorial which can be written as 12 into 11 into 10 factorial upon 2 factorial which is equal to 2 into 1 that is 2 into 10 factorial whole upon 51 factorial which can be written as 51 into 50 into 49 factorial upon 2 factorial that is 2 into 1 which is equal to 2 into 49 factorial so we get 12 into 11 upon 51 into 50 which is equal to 12 into 11 is 132 upon 51 into 50 is equal to 2550 so probability of event A when event E1 has already occurred is equal to 132 upon 2550 similarly probability of event A when event E2 has already occurred is equal to probability of drawing 2 diamond cards given that the lost card is not diamond as the lost card is not diamond so the 2 cards are drawn from 13 diamond cards so we have the probability equal to 13C2 upon 51C2 now by using this formula we have N is equal to 13 and R is equal to 2 so 13C2 can be written as 13 factorial by 2 factorial into 13 minus 2 factorial that is 11 factorial similarly in 51C2 by using this formula we have N is equal to 51 and R is equal to 2 51C2 is equal to 51 factorial upon 2 factorial into 51 minus 2 factorial that is 39 factorial now 13 factorial can be written as 14 into 12 into 11 factorial upon 2 factorial into 11 factorial whole upon 51 factorial can be written as 51 into 50 into 39 factorial upon 2 factorial into 49 factorial which is equal to 13 into 12 upon 51 into 50 which is equal to 156 upon 2550 therefore probability of event A when event E2 has already occurred is equal to 156 upon 2550 and now we have to find the probability of the lost card being the diamond so probability of the lost card being the diamond provided 2 diamond cards are drawn is equal to probability of E1 when A has already occurred hence by using this theorem we have 3 events E1 and E2 such that E1 and E2 are mutually exclusive since E1 intersection E2 is empty and collectively exhaustive as E1 union E2 constitutes the sample space F and we have another event A of drawing 2 diamond cards from the remaining cards and we know that probability of event A is not equal to 0 so we can write probability of E1 when event A has already occurred is equal to probability of E1 into probability of event A when E1 has already occurred whole upon probability of E1 into probability of event A when E1 has already occurred plus probability of E2 into probability of event A when E2 has already occurred and we know that probability of E1 is given by 1 by 4 probability of E2 is given by 3 by 4 probability of event A when E1 has already occurred is equal to 132 upon 2550 and probability of event A when E2 has already occurred is equal to 156 upon 2550 on substituting these values we have probability of E1 that is 1 by 4 into probability of event A when E1 has already occurred that is 132 upon 2550 whole upon probability of E1 that is 1 by 4 into probability of event A when E1 has already occurred that is 132 upon 2550 plus probability of E2 that is 3 by 4 into probability of event A when E2 has already occurred which is given by 156 upon 2550 and further solving we get 33 upon 2550 whole upon 43 plus 39 into 3 that is 117 upon 2550 which is equal to 33 upon 150 that is 11 upon 50 so the probability of the last card being abandoned is given by 11 by 50 which is our required answer this completes our session hope you enjoyed this session