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

[VJesus12]

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.

[EconomistGMATTutor]

hi vjesus12.

Let's take a look at your question.

You can not use the equation (52C1)*(3C1) because you can not choose the cards. You just have to pick them from the top.

Hence first, you pick the top card. Then, there are 51 cards and there are only 3 cards that can form a pair with the first card. Hence, the probability is $$P=\frac{3}{51}=\frac{1}{17}.$$ This is why the correct option is C.

I hope this answer may help you.

Feel free to ask me again if you have any doubt.

Regards.

[Sun Light]

Though i'm not an expert, i think i can help !!!

Imagine you are picking a card out of 52 cards, keeping in mind that you want a pair in the second draw.

At first draw you can simply pick any.
So picking the first card is not related to the chance of picking a pair.
To summarize: your first pick is simply a pick.

Now, the entire maths is in the second pick. Because after the first pic, you know your card. Now, the challenge is to find a pair of that particular card.
Let Ace be that first card, you know that the to pair the Ace you have only 3 Ace cards left out of 51
So probability of picking a pair is 3/51 = 1/17

[Brent@GMATPrepNow]

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

[[email protected]]

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

[Scott@TargetTestPrep]

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

[Scott@TargetTestPrep]

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