1. ## Counting

What is the counting logic behind this question?

2. ## Re: Counting

There are 52 cards in the pack, 4 kings (K) and 48 non-kings (N). There are 3 ways to get "exactly one king" in three cards: "KNN", "NKN", and "NNK". The probability the first card dealt is a king is 4/52. There are then 51 cards left in the deck, 3 kings and 48 non-kings. The probability the second card dealt is a non-king is 48/51. There are then 50 cards, 3 kings and 47 non-kings. The probability the third card dealt is a non-king is 3/50. The probability of "king, non-king, non-king" (KNN) in that order is (4/52)(48/51)(47/50). I will leave it to you to show that the probabilities of "non-king, king, non-king" (NKN) and "non-king, non-king, king" (NNK) are also (4/52)(48/51)(47/50) so that the probability of "exactly one king" in any order is (4/52)(48/51)(47/50)+ (4/52)(48/51)(47/50)+ (4/52)(48/51)(47/50)= 3(4/52)(48/51)(47/50). You can, of course, reduce those fractions.

3. ## Re: Counting

Originally Posted by lionraja

What is the counting logic behind this question?
Look at this calculation, from which can you see the logic?