What is the probability of picking a queen and a jack in two draws fro

Question

What is the probability of picking a queen and a jack in two draws from a pack of 52 cards without replacing any card into the deck after picking it.

A)16/2652
B) 32/2652
C) 16/2704
D) 32/2704
E) 8/2652

CounterSniper · Answer

since there are two draws 
we can pick the card in 2 ways 
queen, jack or jack,queen 
so,
2( \frac{4}{52} ) * ( \frac{4}{51} ) = \frac{32}{2652} 
B

BrentGMATPrepNow · Answer

Since people often post questions involving playing cards on this site, it's worth noting that the GMAT does not expect students to know the composition of a deck of cards (e.g., there are 52 cards in total, 4 suits, etc.). The reason is that questions involving playing cards would be culturally biased, since not all countries allow their citizens to own playing cards.

IF the test makers where to create a question involving a deck of playing cards, the question would undoubtedly explain the composition of a deck of cards first.

IF  the test makers where to create a question involving a deck of playing cards, the question would undoubtedly explain the composition of a deck of cards first.

IanStewart · Answer

Brent is completely right - you don't need to know a thing about playing cards for the GMAT. This particular question is probably of more interest to poker players than to GMAT test takers (at least in poker variants like hold'em where you start with two cards). If you want to know the probability of getting any specific unpaired hand, like queen-jack or ace-three, then, using queen-jack as in the question, our first card can be any of the 8 queens or jacks, so the probability is 8/52 that our first card is one of the ones we want, and then there will be 4 cards left that complete our queen-jack hand, and we'll be picking our second card from 51, so the probability our second card is what we want is 4/51, so the answer is 8/52 * 4/51 = 8/(51*13) = 8/663 ~ 1.2%.

Or you could look at things this way: the probability you get a pair when dealt two cards is 3/51 (once we know our first card, 3 out of the 51 remaining cards will pair it), or 1/17. So the probability we do not get a pair is 16/17. Each specific unpaired hand is equally likely (you're as likely to get ten-three as ace-king), and there are 13C2 = (13*12)/2! = 13*6 different unpaired hands that are possible, so the probability we get some specific unpaired hand like queen-jack is (16/17)*(1/13C2) = 16/(17*13*6) = 8/663.

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What is the probability of picking a queen and a jack in two draws fro

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What is the probability of picking a queen and a jack in two draws fro

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