What is the probability that you get a pair . . . .

VJesus12 wrote:What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards (52 total cards, with 13 sets of four matching suits)? $$(A)\ \frac{12}{2,652}$$ $$(B)\ \frac{16}{2,652}$$ $$(C)\ \frac{1}{17}$$ $$(D)\ \frac{1}{13}$$ $$(E)\ \frac{1}{2}$$ The OA is the option C.

P(select pair) = P(1st card is ANY card AND 2nd card matches 1st card)
= P(1st card is ANY card) x P(2nd card matches 1st card)
= 1 x 3/51
= 3/51
= 1/17
= C

Aside: P(2nd card matches 1st card) = 3/51, because once 1 card is selected, there are 51 cards remaining in the deck. Among those 51 remaining cards, there are 3 that match the 1st card selected.

Cheers,
Brent

Hi VJesus12,

With a standard 52-card deck, each 'value' appears 4 times (four kings, four 10s, four 3s, etc.). The probability of getting a matching pair (meaning ANY pair) is different from the probability of getting a specific pair (for example, two kings). Both of those probabilities appear in the answer choices though, so you have to be careful with these types of questions - make sure that you answer the question that is ASKED.

Since we're looking for the probability of pulling ANY matching pair - and each value appears 4 times - there's an equal probability that the first card will be any of the 13 possibilities. Thus, the first card doesn't really matter - it's the SECOND card that matters (since we need it to match the first). Once that first card is drawn, there are only 3 more cards in the deck (out of 51 remaining cards) that would match the first card. Thus, the probability is 3/51 = 1/17.

Final Answer: C

As an aside, the probability of pulling a SPECIFIC pair (for example, two kings) requires that BOTH cards match the specific value that you're looking for:
For the 1st card, there is a 4/52 probability of drawing that card. Assuming that 1st card matches....
For the 2nd card, there is a 3/51 probability of drawing that card.
Thus, the probability would be (4/52)(3/51) = 12/2652.... but that is the answer to a DIFFERENT question.

GMAT assassins aren't born, they're made,
Rich

VJesus12 wrote:What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards (52 total cards, with 13 sets of four matching suits)? $$(A)\ \frac{12}{2,652}$$ $$(B)\ \frac{16}{2,652}$$ $$(C)\ \frac{1}{17}$$ $$(D)\ \frac{1}{13}$$ $$(E)\ \frac{1}{2}$$ The OA is the option C.

How should I solve this PS question? the solution is (52C1)*(3C1)? Experts, I need your help here. Please.

Suppose the first card has been chosen. Now, among the remaining 51 cards, there are 3 cards that form a pair with the already selected card, and thus the probability of getting a pair is 3/51 = 1/17.

Answer: C

VJesus12 wrote:What is the probability that you get a pair when picking the top two cards off of a randomly shuffled deck of cards (52 total cards, with 13 sets of four matching suits)? $$(A)\ \frac{12}{2,652}$$ $$(B)\ \frac{16}{2,652}$$ $$(C)\ \frac{1}{17}$$ $$(D)\ \frac{1}{13}$$ $$(E)\ \frac{1}{2}$$ The OA is the option C.

How should I solve this PS question? the solution is (52C1)*(3C1)? Experts, I need your help here. Please.

Suppose the first card has been chosen. Now, among the remaining 51 cards, there are 3 cards that form a pair with the already selected card, and thus the probability of getting a pair is 3/51 = 1/17.

Answer: C