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

Question

From an ordinary deck of cards, only 12 picture cards are retained. They are shuffled (mixed) and a man draws2 cards at random. He then draws 2 cards, and announces that he holds at least one King. What is the probability that he holds two Kings in his hand?

A. 3/4
B. 6/13
C. 3/11
D. 3/16
E. 3/19

A. 3/4
B. 6/13
C. 3/11
D. 3/16
E. 3/19

Are You Up For the Challenge: 700 Level Questions: 700 Level Questions

CrackverbalGMAT · Answer

12 picture cards has 4 Kings, 4 Queens and 4 Jacks in it.

P(King) = 4/12 = 1/3 and P(Other) = 1 - 1/3 = 2/3

To Find the probability that he is holding 2 kings given that he is holding at least 1 king

By Bayes Theorem, P(A/B) = P(A given B) = P(A and B) / P(B)

Here P(A and B) = P(hold 2 king and at least 1 king) = P(holding 2 Kings) and P(B) = P(Holding at least 1 king)

P(Holding 2 Kings)

P(K, O) =  \frac{4C1 * 8C1}{12C2} = \frac{4 * 8}{\frac{12 * 11}{2}} = 16/33

P(K, K) =  \frac{4C2}{12C2} = \frac{\frac{4 * 3}{2}}{\frac{12 * 11}{2}} = 1/11

The probability of holding 2 kings = (16/33 + 1/11) = 19/33

The required Probability =  \frac{\frac{1}{11}}{\frac{19}{33}}  = 3/19

Option E

Arun Kumar

ScottTargetTestPrep · Answer

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

Regor60 · Answer

Question needs to be reworded since it mentions drawing two cards twice and the stipulated answer relies on one drawing.

If there is only 1 King selected and a different face card as the other card there are then:

4 Kings * 8 Jacks or Queens =

32 ways one King is selected

If two Kings are selected there are:

4!/2!2! = 6 ways

So the total ways at least one King can be selected is:

32+6 = 38 ways

So the probability of selecting 2 Kings given at least 1 King is:

6/38 = 3/19

gioguidi · Answer

Indeed the question is misleading and worded wrongly. I would solve it as follows:

By Bayes Theorem, P(A/B) = P(A given B) = P(A and B)/ P(B) where:
P(A) probability of drawing 2 kings
P(B) probability of drawing at least 1 king

Please note that P(A) is a subset of P(B), thus P(A and B)=P(A). Therefore:
P(A) = 4/12*3/11 = 3/33 = 1/11
P(B) = 1 - P(0 kings) = 1- (8/12*7/11) = 19/33

P(A)/ P(B) = 3/33 / 19/33 =  3/19 , therefore  option E

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia