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

It is currently 18 Oct 2018, 06:48

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Retired Moderator
User avatar
S
Joined: 18 Sep 2014
Posts: 1132
Location: India
GMAT ToolKit User Premium Member Reviews Badge
What is the probability that you get a pair when picking the top two c  [#permalink]

Show Tags

New post 08 Jun 2016, 10:37
9
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

56% (01:45) correct 44% (01:53) wrong based on 139 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)?
Both give same answer.
SC Moderator
User avatar
D
Joined: 13 Apr 2015
Posts: 1693
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
GPA: 4
WE: Analyst (Retail)
GMAT ToolKit User Premium Member
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

Show Tags

New post 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

Answer: C
Manager
Manager
avatar
Joined: 01 Mar 2014
Posts: 117
Schools: Tepper '18
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

Show Tags

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.!
CEO
CEO
User avatar
D
Joined: 12 Sep 2015
Posts: 3011
Location: Canada
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

Show Tags

New post 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
_________________

Brent Hanneson – GMATPrepNow.com
Image
Sign up for our free Question of the Day emails

Intern
Intern
avatar
B
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

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
Manager
Manager
User avatar
B
Status: IF YOU CAN DREAM IT, YOU CAN DO IT
Joined: 03 Jul 2017
Posts: 207
Location: India
Concentration: Finance, International Business
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

Show Tags

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 ??
Director
Director
User avatar
P
Joined: 27 May 2012
Posts: 583
Premium Member
What is the probability that you get a pair when picking the top two c  [#permalink]

Show Tags

New post 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
User avatar
P
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3863
Location: United States (CA)
Re: What is the probability that you get a pair when picking the top two c  [#permalink]

Show Tags

New post 14 Sep 2018, 17:51
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.

Answer: C
_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

GMAT Club Bot
Re: What is the probability that you get a pair when picking the top two c &nbs [#permalink] 14 Sep 2018, 17:51
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.