Find all School-related info fast with the new School-Specific MBA Forum

It is currently 24 Jul 2014, 05:05

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Four cards are randomly drawn from a pack of 52 cards. Find

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Director
Director
avatar
Joined: 23 Apr 2010
Posts: 584
Followers: 2

Kudos [?]: 23 [0], given: 7

Four cards are randomly drawn from a pack of 52 cards. Find [#permalink] New post 01 Jun 2010, 23:36
1
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (02:02) correct 0% (00:00) wrong based on 3 sessions
Four cards are randomly drawn from a pack of 52 cards. Find the probability that all four cards are of different denominations?

My solution:

[Reveal] Spoiler:
OMEGA = 52 x 51 x 50 x 49
A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68


Official solution:

[Reveal] Spoiler:
OMEGA: (52 4)
A: (4 1) x (13 4)
P (A) = 0.01


I think the official solution is wrong.
Director
Director
avatar
Joined: 23 Apr 2010
Posts: 584
Followers: 2

Kudos [?]: 23 [0], given: 7

Re: Probability cards [#permalink] New post 02 Jun 2010, 00:35
nonameee wrote:
My solution:

[Reveal] Spoiler:
OMEGA = 52 x 51 x 50 x 49
A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68




I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

Your opinion would be greatly appreciated.
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18716
Followers: 3238

Kudos [?]: 22304 [1] , given: 2613

Re: Probability cards [#permalink] New post 02 Jun 2010, 07:44
1
This post received
KUDOS
Expert's post
nonameee wrote:
Four cards are randomly drawn from a pack of 52 cards. Find the probability that all four cards are of different denominations?

My solution:

[Reveal] Spoiler:
OMEGA = 52 x 51 x 50 x 49
A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68


Official solution:

[Reveal] Spoiler:
OMEGA: (52 4)
A: (4 1) x (13 4)
P (A) = 0.01


I think the official solution is wrong.


nonameee wrote:
nonameee wrote:
My solution:

[Reveal] Spoiler:
OMEGA = 52 x 51 x 50 x 49
A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68




I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

Your opinion would be greatly appreciated.


I think you are right:

\frac{4^4*C^4_{13}}{C^4_{52}} would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -C^4_{13};
Multiplying by 4^4 as each card from the above string can be of four suits (4*4*4*4=4^4);
Total # of ways to choose 4 cards out of 52 - C^4_{52}.

Or another way:

We can choose any card for the first one - \frac{52}{52};
Next card can be any card but 3 of the denomination we'v already chosen - \frac{48}{51} (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left);
Next card can be any card but 3*2=6 of the denominations we'v already chosen - \frac{44}{50} (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left);
Last card can be any card but 3*3=9 of the denominations we'v already chosen - \frac{40}{49};

P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49} - the same answer as above.

Hope it helps.


!
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Director
Director
avatar
Joined: 23 Apr 2010
Posts: 584
Followers: 2

Kudos [?]: 23 [0], given: 7

Re: Probability cards [#permalink] New post 02 Jun 2010, 23:41
Bunuel, thank you for your answer.

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. T


Wanted just to correct myself. The above argument is wrong since: 52 x 48 x 44 x 40 is a case of variations without repetition. Therefore, the set won't contain cards of the same denomination.
Manager
Manager
User avatar
Joined: 03 Feb 2010
Posts: 181
Concentration: Strategy
Schools: Northwestern (Kellogg) - Class of 2014
WE: Marketing (Computer Hardware)
Followers: 5

Kudos [?]: 25 [0], given: 34

GMAT Tests User
Re: Probability cards [#permalink] New post 04 Jun 2010, 11:18
nonameee wrote:
Bunuel, thank you for your answer.

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. T


Wanted just to correct myself. The above argument is wrong since: 52 x 48 x 44 x 40 is a case of variations without repetition. Therefore, the set won't contain cards of the same denomination.


Why wouldn't the probability be 1 * 13/51 * 13/50 * 13/49 ? We know there are 4 suits, and the probability of getting any card in one suit is 13/# of cards in the deck.
Manager
Manager
User avatar
Joined: 03 Feb 2010
Posts: 181
Concentration: Strategy
Schools: Northwestern (Kellogg) - Class of 2014
WE: Marketing (Computer Hardware)
Followers: 5

Kudos [?]: 25 [0], given: 34

GMAT Tests User
Re: Probability cards [#permalink] New post 04 Jun 2010, 11:33
Bunuel wrote:
nonameee wrote:
Four cards are randomly drawn from a pack of 52 cards. Find the probability that all four cards are of different denominations?

My solution:

[Reveal] Spoiler:
OMEGA = 52 x 51 x 50 x 49
A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68


Official solution:

[Reveal] Spoiler:
OMEGA: (52 4)
A: (4 1) x (13 4)
P (A) = 0.01


I think the official solution is wrong.


nonameee wrote:
nonameee wrote:
My solution:

[Reveal] Spoiler:
OMEGA = 52 x 51 x 50 x 49
A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68




I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

Your opinion would be greatly appreciated.


I think you are right:

\frac{4^4*C^4_{13}}{C^4_{52}} would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -C^4_{13};
Multiplying by 4^4 as each card from the above string can be of four suits (4*4*4*4=4^4);
Total # of ways to choose 4 cards out of 52 - C^4_{52}.

Or another way:

We can choose any card for the first one - \frac{52}{52};
Next card can be any card but 3 of the denomination we'v already chosen - \frac{48}{51} (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left);
Next card can be any card but 3*2=6 of the denominations we'v already chosen - \frac{44}{50} (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left);
Last card can be any card but 3*3=9 of the denominations we'v already chosen - \frac{40}{49};

P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49} - the same answer as above.

Hope it helps.

No posting of PS/DS questions is allowed in the main Math forum.[/warning]


In your solution, you found the # of ways to choose 4 different denominations out of 13, I believe that is incorrect. Each suit(denomination) has 13 cards, we're looking for the # of ways to draw 1 card of a different suit each time, up to 4 draws. Correct me if I'm wrong.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18716
Followers: 3238

Kudos [?]: 22304 [0], given: 2613

Re: Probability cards [#permalink] New post 04 Jun 2010, 12:09
Expert's post
edoy56 wrote:
nonameee wrote:
Bunuel, thank you for your answer.

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. T


Wanted just to correct myself. The above argument is wrong since: 52 x 48 x 44 x 40 is a case of variations without repetition. Therefore, the set won't contain cards of the same denomination.


Why wouldn't the probability be 1 * 13/51 * 13/50 * 13/49 ? We know there are 4 suits, and the probability of getting any card in one suit is 13/# of cards in the deck.


edoy56 wrote:
In your solution, you found the # of ways to choose 4 different denominations out of 13, I believe that is incorrect. Each suit(denomination) has 13 cards, we're looking for the # of ways to draw 1 card of a different suit each time, up to 4 draws. Correct me if I'm wrong.


Yes you are wrong.

First of all suit and denomination are not the same: there are four suits (Clubs, Diamonds, Hearts, and Spades), and 13 different denominations (Ace, two through ten, Jack, Queen, and King).

Next, even if you are counting the probability that the 4 cards you draw are of different suits, 1 * 13/51 * 13/50 * 13/49 won't be the correct answer.

Correct answer would be P=\frac{(C^1_{13})^4}{C^4_{52}}.

Or another way: P=4!*\frac{13}{52}*\frac{13}{51}*\frac{13}{50}*\frac{13}{49} (the same probability as above), we are multiplying by 4! as the favorable scenario of Clubs, Diamonds, Hearts, and Spades (CDHS) can occur in 4! # of ways, which is basically the # of permutaions of 4 letters CDHS.

Hope it helps.

P.S. This question is beyond the scope of GMAT, so I don't recommend to spend too much time on it.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 03 Feb 2010
Posts: 181
Concentration: Strategy
Schools: Northwestern (Kellogg) - Class of 2014
WE: Marketing (Computer Hardware)
Followers: 5

Kudos [?]: 25 [0], given: 34

GMAT Tests User
Re: Probability cards [#permalink] New post 04 Jun 2010, 12:21
Thanks for the explanation, makes perfect sense now
Intern
Intern
avatar
Joined: 29 Dec 2009
Posts: 33
Followers: 0

Kudos [?]: 2 [0], given: 2

Re: Probability cards [#permalink] New post 07 Jun 2010, 10:34
confused :(

why is it multiplied by 4^4 and not 4!?
if the first card is taken from 1 suit the second can be from the remaining 3 because 1 suit will have only one card of one denomination....?
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18716
Followers: 3238

Kudos [?]: 22304 [1] , given: 2613

Re: Probability cards [#permalink] New post 07 Jun 2010, 10:53
1
This post received
KUDOS
Expert's post
ssgomz wrote:
confused :(

why is it multiplied by 4^4 and not 4!?
if the first card is taken from 1 suit the second can be from the remaining 3 because 1 suit will have only one card of one denomination....?


We need 4 cards of different denomination, but these 4 cards don't have to be of the different suits.

Consider this: C^4_{13} is the # of ways to choose 4 cards of different denomination. For example one particular case from C^4_{13} would be: Jack - Queen - King - Ace.

Now, each card from this case (Jack - Queen - King - Ace) can be of 4 suits: Clubs, Diamonds, Hearts, and Spades. So taking into account the suits the case Jack - Queen - King - Ace can be formed 4*4*4*4=4^4 # of times. As there are C^4_{13} # of cases like Jack - Queen - King - Ace, so total # of ways to form 4 cards of different denomination is C^4_{13}*4^4.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 29 Dec 2009
Posts: 33
Followers: 0

Kudos [?]: 2 [0], given: 2

Re: Probability cards [#permalink] New post 07 Jun 2010, 11:05
Thank you for the detailed explanation. It is clear now :)
Intern
Intern
avatar
Joined: 28 May 2010
Posts: 4
Followers: 0

Kudos [?]: 2 [0], given: 1

Re: Probability cards [#permalink] New post 09 Jun 2010, 09:24
Bunuel wrote:
nonameee wrote:
Four cards are randomly drawn from a pack of 52 cards. Find the probability that all four cards are of different denominations?

My solution:

[Reveal] Spoiler:
OMEGA = 52 x 51 x 50 x 49
A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68


Official solution:

[Reveal] Spoiler:
OMEGA: (52 4)
A: (4 1) x (13 4)
P (A) = 0.01


I think the official solution is wrong.


nonameee wrote:
nonameee wrote:
My solution:

[Reveal] Spoiler:
OMEGA = 52 x 51 x 50 x 49
A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68




I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

Your opinion would be greatly appreciated.


I think you are right:

\frac{4^4*C^4_{13}}{C^4_{52}} would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -C^4_{13};
Multiplying by 4^4 as each card from the above string can be of four suits (4*4*4*4=4^4);
Total # of ways to choose 4 cards out of 52 - C^4_{52}.

Or another way:

We can choose any card for the first one - \frac{52}{52};
Next card can be any card but 3 of the denomination we'v already chosen - \frac{48}{51} (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left);
Next card can be any card but 3*2=6 of the denominations we'v already chosen - \frac{44}{50} (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left);
Last card can be any card but 3*3=9 of the denominations we'v already chosen - \frac{40}{49};

P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49} - the same answer as above.


How did \frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49} become equal to \frac{4^4*13*12*11*10}{52*51*50*49} here?

Take 4 common from the numerator and the result would be 4*\frac{13*12*11*10}{52*51*50*49}.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18716
Followers: 3238

Kudos [?]: 22304 [0], given: 2613

Re: Probability cards [#permalink] New post 09 Jun 2010, 09:32
Expert's post
narisipalli wrote:
How did \frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49} become equal to \frac{4^4*13*12*11*10}{52*51*50*49} here?

Take 4 common from the numerator and the result would be 4*\frac{13*12*11*10}{52*51*50*49}.


Hi,

\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4*13}{52}*\frac{4*12}{51}*\frac{4*11}{50}*\frac{4*10}{49}=\frac{4^4*13*12*11*10}{52*51*50*49}.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 07 Jun 2010
Posts: 50
Location: United States
Followers: 1

Kudos [?]: 8 [0], given: 7

Re: Probability cards [#permalink] New post 12 Jun 2010, 21:56
I think we we should not multiply it with 4 since we are not finding number of ways just probability so order doesnot matter..

13/52 * 13/51*13/49*13/48 should be the answer ...

Please correct me if i am wrong.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18716
Followers: 3238

Kudos [?]: 22304 [0], given: 2613

Re: Probability cards [#permalink] New post 13 Jun 2010, 02:18
Expert's post
virupaksh2010 wrote:
I think we we should not multiply it with 4 since we are not finding number of ways just probability so order doesnot matter..

13/52 * 13/51*13/49*13/48 should be the answer ...

Please correct me if i am wrong.


Your approach is not correct:
First of all Probability \ of \ an \ event =\frac {# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}, so we do need to find the # of ways and if order matters for # of ways of an event, then order matters for probability too.

Answer for the original question is C^4_{52}=\frac{4^4*13*12*11*10}{52*51*50*49} ( so we are not multiplying 13/52 * 13/51*13/49*13/48 by 4 we are multiplying by 4^4), please see the solution in my first post.

If you are referring to the different question we've considered (the probability that the 4 cards you draw are of different suits), the answer for it is still not 4*13/52 * 13/51*13/49*13/48 it's P=\frac{(C^1_{13})^4}{C^4_{52}}=4!*\frac{13}{52}*\frac{13}{51}*\frac{13}{50}*\frac{13}{49}, so again we are not multiplying by 4 we are multiplying by 4!=1*2*3*4=24, please see the solution for this question in my second post.


For more on this issue please check Probability and Combinatorics chapters of Math Book (link in my signature).

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: Probability cards   [#permalink] 13 Jun 2010, 02:18
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic Two cards are drawn successively from a standard deck of 52 GMATMadeeasy 8 08 Jan 2010, 16:29
20 cards are numbered 1-20 and drawn randomly from a hat and crankharder 9 10 Feb 2007, 14:32
1) Cards are drawn (without replacement) from a deck (of 52 bigpapi 6 27 Mar 2006, 17:50
In how many ways can 3 cards be selected from a pack of 52 sandman 1 26 Nov 2005, 12:17
Two cards are drawn at random from a pack of 52 cards. Find remgeo 6 22 Nov 2005, 06:26
Display posts from previous: Sort by

Four cards are randomly drawn from a pack of 52 cards. Find

  Question banks Downloads My Bookmarks Reviews Important topics  


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.