A card from a pack of 52 cards is lost. From 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?

The number of ways to choose 2 diamonds cards if the lost card was a diamond = 12C2/51C2. The number of ways to choose 2 diamonds cards if the lost card was not a diamond = 3* 13C2/51C2. probability of getting any one suit is 1/4 thus using Bayes' theorem 1/4 * [12C2/51C2]/ [1/4 * [12C2/51C2] + 1/4 *3* [13C2/51C2]] Cancelling 1/4 and 51C2 from numerator and denominator = 12C2/ [12C2+(3 * 13C2)] = 66/300 = 11/50 read less

Let E1 be the event that the lost card is a diamond. Given that there are 13 diamonds in the deck, P(E1) = 13/52=14 Let E2 be the event that the lost card is not a diamond. P(E2) =1 - 13/52=39/52=34 Let A be the event that the two cards drawn are found to be both diamonds. We need to calculate P (2 cards are both diamond w the lost card being a diamond) next. Given 13 diamond cards, if one is lost, then we have 12 remaining diamonds out of 51 total cards now. The two cards that are both diamond can be drawn in 12C2=12*11/1*2=13*22 = 66 ways. Likewise, the two diamonds can be drawn from the pack of 51 remaining cards in 51C2=51*50/1*2=25502 = 1275 ways. Therefore, the P (2 cards drwan are diamond given one is lost) = P (A|E1) = 66/1275 Now, consider the event where the two cards drawn are both diamonds but the lost card is not a diamond. The two cards that are both diamond can be drawn in 13C2=13*12/1*2=1562 = 78ways. Likewise, the two diamonds can be drawn from the pack of 51 remaining cards in 51C2=51*50/1*2=2550/2 = 1275 ways. Therefore, the P (2 cards drwan are diamond given one card which is not a diamond is lost) = P (A|E1) = 78/1275 We need to calcuate the probability that the lost card is a diamond. P (E1|A). We can use Baye's theorem, according to which P(E1|A)=P(E1)(P(A|E1)/P(E1)P(A|E1)+P(E2)P(A|E2) P(E1|A) = ((66/1275).(1/4))/((66/1275).(1/4))+((78/1275).(3/4))=66/(66+78*3)=66/300=11/50 read less