From an ordinary deck of cards, only 12 picture cards are retained. Th

Solution:

Let’s first look at the 4 different outcomes that can happen when 2 cards are drawn from this deck of 4 Jacks, 4 Queens, and 4 Kings. Letting K = the outcome that a King is drawn and N = the outcome that a King is not drawn, we have:

(K, K) or (K, N) or (N, K) or (N, N)

Let’s calculate the probability of each outcome:

P(K, K) = 4/12 x 3/11 = 12/132 = 3/33 = 1/11

P(K, N) = 4/12 x 8/11 = 32/132 = 8/33

P(N, K) = 8/12 x 4/11 = 32/132 = 8/33

P(N, N) = 8/12 x 7/11 = 56/132 = 14/33

We see that the event “at least one King is drawn” includes the first 3 outcomes above: (K,K), (K,N), and (N,K), and the total probability that one of these 3 outcomes will happen is the sum of their individual probabilities, which is 1/11 + 8/33 + 8/33 = 3/33 + 8/33 + 8/33 = 19/33.

But, out of those three outcomes, only (K, K), with probability 3/33, is relevant to answering the question “what is the probability that he has two Kings in his hand?” Thus, the probability that he has two Kings in his hand, given that he has at least one King, is (3/33) / 19/33) = 3/19.

Answer: E