 Hello and welcome to the session. In this session we discussed the following question which says a card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be both clubs. Find the probability of the lost card being of clubs. Before that, let's discuss the conditional probability and base theorem. First we have the conditional probability of the event E given that is given by number of events favorable to should F upon the number of events favorable to the event. That is, we have that is the conditional probability of the event E given that F has already occurred is equal to probability of E intersection F is not equal to 0. Next we have the base theorem disjoint and our sample space S is equal to even union E to union so n union E n equal to 0. Then we say the probability of the conditional probability of the event E i given that event A has already occurred is equal to probability of the event E i into the conditional probability of the event A given that i has already occurred probability of the event E j into probability of the event A given that E j has already occurred. This j goes from 1 to n and also we have i equal to 1, 2, 3 and so on a t idea that we use for this question. Let's see with the solution now the question we have that we have a pack of 52 cards and one card is lost from this pack of 52 cards and from the remaining cards which are left find the probability of the lost card being lost. Let's define some events consider event even the event such that that the lost card which is the event find the probability of the lost card being of clubs. This means we need to find the conditional probability of the event E3 such that event A has already occurred. That is find the probability that the lost card of a class given that using this formula we can find out the probability that event A has already occurred would be equal to probability of event E3 into probability of event A given that event E3 has already occurred. This upon probability of event E1 into probability of event has already occurred plus the probability of event E2 into probability of event A given that E2 has already occurred plus probability of event E3 into probability of event A given that E3 has already occurred the probability of event E4 into probability of event A given that E4 has already occurred. Now given that E3 has already occurred is the of clubs given that we have 13 club cards and from 13 club cards we lost one club card. So we are left with 12 club cards and from there we choose 2 cards. So we have 12 C2 upon 51 C2, cards 1 card is lost so we left with 50 out the probability of event A given that even has already occurred that is the probability of going 2 club cards given that card is lost would be equal to 13 C2 so there are 13 hard cards and from there we choose 2 cards upon 51 C2 where 51 are the remaining 1 card is lost and from there we choose 2 cards so this is the probability of event A given that even has already occurred next is the probability of event A given that event E2 have already occurred is the probability of drawing 2 club cards given that once paid card is lost these paid cards and from there we choose 2 cards so 13 51 C2 then next we have probability of the event A given that event E4 is already occurred which is the probability of drawing 2 club cards given that the lost card is of diamonds would be equal to 13 C2 so there are 13 diamond cards upon 51 C2 find out the probability of event E1 card is of hearts as there are 13 cards of hearts here even is equal to 13 upon the total number of cards in the pack which is 52 where 13 4 times is 52 so this is equal to 1 upon 4 in the same way probability of E2 even that the lost card is of space as we know there are 13 cards of so 13 upon 52 upon 4 then the probability of the event E3 even that the lost card would be equal to 13 since there are 13 clubs in the pack of cards so 13 upon 52 then probability of equal to cards of diamonds so 13 upon 52 then upon that we have to put the values in this formula with this we the values 3 given that event A has already occurred is equal to probability of E3 which is 1 upon 4 F event A given that E3 has already occurred is equal to 12 C so here we have 1 upon 4 into C2 upon 15 and even that is 1 upon 4 into the probability of the event A given that even has already occurred is 13 C2 upon C right here then upon 4 into 13 C2 upon 51 C2 probability of E2 which is 1 upon 4 into E2 has already occurred which is 13 C2 upon 50 so here we have 1 upon 4 into 13 C2 upon 51 C2 which is the probability of the event E3 into which is 12 C2 upon so here we write C2 upon 51 C2 E4 which is 1 upon 4 into here event A given that E4 has already occurred is 13 C2 upon 51 so we write here 1 upon 4 into C2 equals to 12 C2 into 1 upon 4 into 1 C2 and this into 13 C2 the event upon 4 51 C2 cancels with 12 C2 and so we are left with 12 C2 into C2 plus 12 would be equal to into 12 minus 2 minus upon 3 into 13 minus 2 minus 11 factorial 2 which is 10 factorial now this would be equal to 12 into 11 13 into 12 upon 1 into 2 plus 12 into 11 2 into 2 now this 1 into 2 1 into 2 1 into 2 cancels with each other so we are left with 12 into 11 upon 2 into 12 plus 12 into 11 equal to 12 into 11 into 13 plus 11 we hold now I have this 12 12 cancels so we get this is equal to 39 plus 11 which is equal to 11 upon 50 E3 given that A has already occurred and this is what we were supposed to find so we have the problem D this completes the session hope you understood the solution of this question