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

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Retired Moderator
Joined: 18 Sep 2014
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What is the probability that you get a pair when picking the top two c  [#permalink]

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08 Jun 2016, 10:37
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58% (01:49) correct 42% (01:59) wrong based on 122 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) $$\frac{1}{2}$$

Pair of same design(set of 13) or matching suits(types of 4)?
Both give same answer.
Marshall & McDonough Moderator
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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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08 Jun 2016, 11:36
1
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

Manager
Joined: 01 Mar 2014
Posts: 108
Schools: Tepper '18
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

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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

Did this -

52C1*3C1 /52C2

It is coming out to 2/17. What am I doing wrong here?
Thanks.!
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Joined: 12 Sep 2015
Posts: 4003
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

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27 Jan 2018, 10:58
Top Contributor
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) $$\frac{1}{2}$$

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.

Cheers,
Brent
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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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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

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
Manager
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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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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 ??
Director
Joined: 27 May 2012
Posts: 902
What is the probability that you get a pair when picking the top two c  [#permalink]

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11 Sep 2018, 06:39
1
longhaul123 wrote:
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 ??

Hi longhaul123,
Hope I am not too late here , anyway.

For the first card the probability is $$\frac{52}{52}$$and NOT $$\frac {1}{52}$$ , because you can select any of the 52 cards out of a total 52 cards.
For the 2nd card we have only 3 options , because there are only 4 of the same number, four 1's , four 2's , .....four kings, four queens and four aces .So for any first number we pick there are only 3 other remaining of the same number.
Hence the probability is $$\frac {52}{52} * \frac{3}{51} = \frac{1}{17}$$

Hope this helps. Let me know if anything is still unclear.
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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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14 Sep 2018, 17:51
1
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) $$\frac{1}{2}$$

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.

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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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12 Feb 2019, 14:01
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) $$\frac{1}{2}$$

$$?\,\, = \,\,P\left( {{\rm{get}}\,\,{\rm{a}}\,\,{\rm{pair}}} \right)$$

$$\left. \matrix{ {\rm{Total}}\,\,:\,\,\,C\left( {52,2} \right) = {{52 \cdot 51} \over 2} = 26 \cdot 51\,\,\,{\rm{equiprobable}}\,\,{\rm{choices}}{\kern 1pt} \,\, \hfill \cr {\rm{Favorable}}\,\,:\,\,\,\underbrace {\,C\left( {4,2} \right)}_{2\,\,{\rm{suits}}} \cdot \underbrace {C\left( {13,1} \right)}_{{\rm{card}}\,{\rm{to}}\,{\rm{get}}\,{\rm{twice}}} = 6 \cdot 13\,\,{\rm{choices}} \hfill \cr} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\, = \,\,{{6 \cdot 13} \over {26 \cdot 51}}\,\, = \,\,{1 \over {17}}$$

The correct answer is (C).

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: What is the probability that you get a pair when picking the top two c   [#permalink] 12 Feb 2019, 14:01
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# What is the probability that you get a pair when picking the top two c

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