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What is the probability that you get a pair when picking the top two c

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What is the probability that you get a pair when picking the top two c [#permalink]

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New post 08 Jun 2016, 10:37
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Question Stats:

56% (01:01) correct 44% (01:21) wrong based on 78 sessions

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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) \(\frac{1}{2}\)

[Reveal] Spoiler:
Pair of same design(set of 13) or matching suits(types of 4)?
Both give same answer.
[Reveal] Spoiler: OA

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Re: What is the probability that you get a pair when picking the top two c [#permalink]

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New post 08 Jun 2016, 11:36
First pick --> Pick any card
Second pick --> Out of the remaining 51 cards, only 3 cards can match the first card

Probability = 3/51 = 1/17

Answer: C
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Re: What is the probability that you get a pair when picking the top two c [#permalink]

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New post 14 Jun 2016, 11:54
Vyshak wrote:
First pick --> Pick any card
Second pick --> Out of the remaining 51 cards, only 3 cards can match the first card

Probability = 3/51 = 1/17

Answer: C


Did this -

52C1*3C1 /52C2

It is coming out to 2/17. What am I doing wrong here?
Thanks.!
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Re: What is the probability that you get a pair when picking the top two c [#permalink]

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New post 27 Jan 2018, 10:58
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Nevernevergiveup 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) \(\frac{1}{2}\)

[Reveal] Spoiler:
Pair of same design(set of 13) or matching suits(types of 4)?
Both give same answer.


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.

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Re: What is the probability that you get a pair when picking the top two c [#permalink]

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New post 31 Jan 2018, 20:18
bethebest wrote:
Vyshak wrote:
First pick --> Pick any card
Second pick --> Out of the remaining 51 cards, only 3 cards can match the first card

Probability = 3/51 = 1/17

Answer: C


Did this -

52C1*3C1 /52C2

It is coming out to 2/17. What am I doing wrong here?
Thanks.!



hello,

I don't think you can use this distribution in this way because the numerator has overlapping sets. specifically, the numerator above says that you're choosing one card out of the 3 cards, and one card out of 52 cards which would includes those other 3 (in other words you don't want these sets to overlap in this type of distribution)

the best way of approaching this problem was mentioned above - no matter what you pick as the first card, there is a 3 / 51 chance of getting a pair (3 cards that will give you a pair out of the remaining 51)

hope this helps
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Re: What is the probability that you get a pair when picking the top two c [#permalink]

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New post 15 May 2018, 09:19
can someone please let me know where am I going wrong.
Now for the first card the probability is 1/52
now there are 3 pairs left in the other 3 sets and there is a total of 51 cards left.
So the probability is 1/52*3/51
=1/52*1/17
why is this method incorrect ??
Re: What is the probability that you get a pair when picking the top two c   [#permalink] 15 May 2018, 09:19
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What is the probability that you get a pair when picking the top two c

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