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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > A card from a pack of 52 cards is lost. From remaining...
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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?

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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 * / + 1/4 *3* ] Cancelling...
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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
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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...
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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
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GMAT Math Expert

If the lost card is diamond, we are left with 12 cards that are diamonds. So 2 cards can be selected out 12 diamond cards in 12C2 ways and 2 cards can be selected from total number of cards in 51C2 ways (one card is lost). Probability = 12C2/51C2
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12C2/51C2
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Data Analytics & Advanced Statistics Tutor with 18 years of experience in teaching

By applying baye's rule,required probability is 11/50.
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12C2/51C2
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