GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Oct 2019, 16:45

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 18 Sep 2014
Posts: 1096
Location: India
What is the probability that you get a pair when picking the top two c  [#permalink]

### Show Tags

08 Jun 2016, 10:37
11
00:00

Difficulty:

55% (hard)

Question Stats:

58% (01:49) correct 42% (01:59) wrong based on 122 sessions

### HideShow timer Statistics

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)?
Marshall & McDonough Moderator
Joined: 13 Apr 2015
Posts: 1686
Location: India
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

### Show Tags

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]

### Show Tags

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.!
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4003
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

### Show Tags

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)?

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
_________________
Test confidently with gmatprepnow.com
Intern
Joined: 07 Oct 2013
Posts: 18
GMAT 1: 770 Q50 V47
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

### Show Tags

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
Status: IF YOU CAN DREAM IT, YOU CAN DO IT
Joined: 03 Jul 2017
Posts: 187
Location: India
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

### Show Tags

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]

### Show Tags

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.
_________________
- Stne
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8048
Location: United States (CA)
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

### Show Tags

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

### Show Tags

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) $$\frac{1}{2}$$

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

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

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

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Re: What is the probability that you get a pair when picking the top two c   [#permalink] 12 Feb 2019, 14:01
Display posts from previous: Sort by