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

It is currently 30 Aug 2014, 00:28

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

If 40 people get the chance to pick a card from a canister

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 10 Jul 2010
Posts: 1
Followers: 0

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

If 40 people get the chance to pick a card from a canister [#permalink] New post 10 Jul 2010, 05:08
1
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

33% (01:30) correct 67% (00:24) wrong based on 4 sessions
If 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win?
Manager
Manager
avatar
Joined: 03 May 2010
Posts: 89
WE 1: 2 yrs - Oilfield Service
Followers: 11

Kudos [?]: 65 [0], given: 7

GMAT Tests User Reviews Badge
Re: probability of picking last [#permalink] New post 10 Jul 2010, 07:17
35 - lose, 5 - win

Pick 5 people to win => 40C5 = total number of outcomes.

Favorable outcome is : First pick the 40th person, then pick any other 4.
=> 1*40C4

So, probability = 40C5 / 40C4

= 40!*36!*4! / (35!*5!*40!)

= 36/(35*5)

= 36 / 175
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25206
Followers: 3419

Kudos [?]: 25052 [0], given: 2702

Re: probability of picking last [#permalink] New post 10 Jul 2010, 07:54
Expert's post
mpevans wrote:
if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win?


I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass?

If so, then probability of picking the pass will be the same for all 40 people - \frac{5}{40}, (initial probability of picking the pass (\frac{5}{40}) will be the same for any person in a row).

AbhayPrasanna wrote:
35 - lose, 5 - win

Pick 5 people to win => 40C5 = total number of outcomes.

Favorable outcome is : First pick the 40th person, then pick any other 4.
=> 1*40C4

So, probability = 40C5 / 40C4

= 40!*36!*4! / (35!*5!*40!)

= 36/(35*5)

= 36 / 175


This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is 1*C^4_{39}. Also probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes} and you wrote vise-versa.

So P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}.

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

Manager
Manager
avatar
Joined: 03 Jun 2010
Posts: 184
Location: United States (MI)
Concentration: Marketing, General Management
WE: Business Development (Consumer Products)
Followers: 4

Kudos [?]: 21 [0], given: 40

GMAT ToolKit User GMAT Tests User
Re: probability of picking last [#permalink] New post 10 Jul 2010, 21:13
I made the same.
4C39/5C40.
Where 5C40 - total # of outcomes.
4C39 means that 4 winning tickets were taken out by 39 persons.
Manager
Manager
avatar
Joined: 11 Jul 2010
Posts: 229
Followers: 1

Kudos [?]: 39 [0], given: 20

GMAT ToolKit User GMAT Tests User
Re: probability of picking last [#permalink] New post 11 Jul 2010, 05:03
The number of passes here is 40 (35 +5)
And the number of people is also 40

How will this problem change if there are 10 passes available and 45 blank passes mixed in and there are 40 people?

will the probability of 40th person picking the pass be 10/55 = 2/11?

Can someone explain the favorable outcomes/total outcomes set-up using combination's formula? thanks.
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25206
Followers: 3419

Kudos [?]: 25052 [1] , given: 2702

Re: probability of picking last [#permalink] New post 11 Jul 2010, 06:41
1
This post received
KUDOS
Expert's post
gmat1011 wrote:
The number of passes here is 40 (35 +5)
And the number of people is also 40

How will this problem change if there are 10 passes available and 45 blank passes mixed in and there are 40 people?

will the probability of 40th person picking the pass be 10/55 = 2/11?

Can someone explain the favorable outcomes/total outcomes set-up using combination's formula? thanks.


Yes, if there are 10 passes and 45 blank cards and only 40 people are to pick the cards the probability that 40th person will pick the pass will still be 10/55.

Consider another example the deck of 52 cards. If we put them in a line, what is the probability that 40th card will be an ace? As there are 4 aces then probability that any particular card in a line is an ace is 4/52.

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

Manager
Manager
avatar
Joined: 26 Mar 2010
Posts: 125
Followers: 2

Kudos [?]: 5 [0], given: 17

Re: probability of picking last [#permalink] New post 12 Jul 2010, 10:41
Bunuel wrote:
mpevans wrote:
if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win?


I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass?

If so, then probability of picking the pass will be the same for all 40 people - \frac{5}{40}, (initial probability of picking the pass (\frac{5}{40}) will be the same for any person in a row).

AbhayPrasanna wrote:
35 - lose, 5 - win

Pick 5 people to win => 40C5 = total number of outcomes.

Favorable outcome is : First pick the 40th person, then pick any other 4.
=> 1*40C4

So, probability = 40C5 / 40C4

= 40!*36!*4! / (35!*5!*40!)

= 36/(35*5)

= 36 / 175


This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is 1*C^4_{39}. Also probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes} and you wrote vise-versa.

So P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}.

Hope it helps.



Hi... thanks for the explanation Bunuel


But I do not understand how can the probability of selecting a free pass =5/40 for all in case we assume the people are picking the cards and keeping it with them.

won't it keep reducing as 4/39 for the second successfull fick of a free card...
Please explain it seems i am missing some logic somewhere.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25206
Followers: 3419

Kudos [?]: 25052 [0], given: 2702

Re: probability of picking last [#permalink] New post 12 Jul 2010, 10:54
Expert's post
utin wrote:
Hi... thanks for the explanation Bunuel

But I do not understand how can the probability of selecting a free pass =5/40 for all in case we assume the people are picking the cards and keeping it with them.

won't it keep reducing as 4/39 for the second successfull fick of a free card...
Please explain it seems i am missing some logic somewhere.


Consider this: put 40 cards on the table and 40 people against them. What is the probability that the card which is against the 40th person is the winning one? 5/40, it's the same probability as for the first, second, ... for any. When we pick the cards from a canister and not knowing the results till the end then it's basically the same scenario.
_________________

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

1 KUDOS received
Intern
Intern
avatar
Joined: 12 Jan 2012
Posts: 20
GMAT 1: 720 Q49 V39
Followers: 0

Kudos [?]: 12 [1] , given: 10

Re: If 40 people get the chance to pick a card from a canister [#permalink] New post 21 Dec 2012, 22:35
1
This post received
KUDOS
Consider 40 places to arrange the 40 cards with 35 blank and 5 passes
= 40!/(35!*5!)
Favorable outcome is when the 40th place contains a pass, so we have 39 places to arrange 35 blanks and 4 passes
= 39!/(4!*35!)
=(39!*35!*5!)/(40!*35!*4!) = 1/8
Intern
Intern
avatar
Joined: 20 Dec 2012
Posts: 19
Followers: 0

Kudos [?]: 2 [0], given: 3

Re: If 40 people get the chance to pick a card from a canister [#permalink] New post 22 Dec 2012, 05:43
geneticsgene wrote:
Consider 40 places to arrange the 40 cards with 35 blank and 5 passes
= 40!/(35!*5!)
Favorable outcome is when the 40th place contains a pass, so we have 39 places to arrange 35 blanks and 4 passes
= 39!/(4!*35!)
=(39!*35!*5!)/(40!*35!*4!) = 1/8


Hello!

I am unfamiliar with ! in math, what does it mean?

Thanks in advance
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25206
Followers: 3419

Kudos [?]: 25052 [1] , given: 2702

Re: If 40 people get the chance to pick a card from a canister [#permalink] New post 22 Dec 2012, 05:52
1
This post received
KUDOS
Expert's post
Hiho wrote:
geneticsgene wrote:
Consider 40 places to arrange the 40 cards with 35 blank and 5 passes
= 40!/(35!*5!)
Favorable outcome is when the 40th place contains a pass, so we have 39 places to arrange 35 blanks and 4 passes
= 39!/(4!*35!)
=(39!*35!*5!)/(40!*35!*4!) = 1/8


Hello!

I am unfamiliar with ! in math, what does it mean?

Thanks in advance


The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n.

For example: 4!=1*2*3*4=24.

For more check here: everything-about-factorials-on-the-gmat-85592.html

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

Intern
Intern
avatar
Joined: 20 Dec 2012
Posts: 19
Followers: 0

Kudos [?]: 2 [0], given: 3

Re: If 40 people get the chance to pick a card from a canister [#permalink] New post 22 Dec 2012, 06:22
Bunuel wrote:
Hiho wrote:
geneticsgene wrote:
Consider 40 places to arrange the 40 cards with 35 blank and 5 passes
= 40!/(35!*5!)
Favorable outcome is when the 40th place contains a pass, so we have 39 places to arrange 35 blanks and 4 passes
= 39!/(4!*35!)
=(39!*35!*5!)/(40!*35!*4!) = 1/8


Hello!

I am unfamiliar with ! in math, what does it mean?

Thanks in advance


The factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n.

For example: 4!=1*2*3*4=24.

Hope it helps.


Yes, it does. Thanks. :)

I understand the concept, but not the use of it in this particular case.

Is it tested on the GMAT, or is it just additional help on some questions for those who are familiar with it?
Intern
Intern
avatar
Joined: 22 Dec 2012
Posts: 16
GMAT 1: 720 Q49 V39
Followers: 1

Kudos [?]: 10 [0], given: 19

Re: probability of picking last [#permalink] New post 22 Dec 2012, 10:38
Hi,

Please help me out here...
we have 40 cards with 5 valid passes and rest junks
we have 40 people ...
The probability of 1st person picking junk is 35/40 and then he doesnt replace the card rite.. he takes it with him.. so now we are left with 39 cards.. The probability of 2nd person taking a junk card is 34/39 right??? so wont it be

35/40 x 34/39 x 33/38 x ...... 1/5???

what am I missing here pls?


Bunuel wrote:
mpevans wrote:
if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win?


I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass?

If so, then probability of picking the pass will be the same for all 40 people - \frac{5}{40}, (initial probability of picking the pass (\frac{5}{40}) will be the same for any person in a row).

AbhayPrasanna wrote:
35 - lose, 5 - win

Pick 5 people to win => 40C5 = total number of outcomes.

Favorable outcome is : First pick the 40th person, then pick any other 4.
=> 1*40C4

So, probability = 40C5 / 40C4

= 40!*36!*4! / (35!*5!*40!)

= 36/(35*5)

= 36 / 175


This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is 1*C^4_{39}. Also probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes} and you wrote vise-versa.

So P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}.

Hope it helps.
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25206
Followers: 3419

Kudos [?]: 25052 [0], given: 2702

Re: If 40 people get the chance to pick a card from a canister [#permalink] New post 23 Dec 2012, 05:18
Expert's post
Hiho wrote:
Yes, it does. Thanks. :)

I understand the concept, but not the use of it in this particular case.

Is it tested on the GMAT, or is it just additional help on some questions for those who are familiar with it?


It is tested.

Check here: math-combinatorics-87345.html and here: math-probability-87244.html
_________________

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

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 25206
Followers: 3419

Kudos [?]: 25052 [0], given: 2702

Re: probability of picking last [#permalink] New post 23 Dec 2012, 05:22
Expert's post
SpotlessMind wrote:
Hi,

Please help me out here...
we have 40 cards with 5 valid passes and rest junks
we have 40 people ...
The probability of 1st person picking junk is 35/40 and then he doesnt replace the card rite.. he takes it with him.. so now we are left with 39 cards.. The probability of 2nd person taking a junk card is 34/39 right??? so wont it be

35/40 x 34/39 x 33/38 x ...... 1/5???

what am I missing here pls?


Bunuel wrote:
mpevans wrote:
if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win?


I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass?

If so, then probability of picking the pass will be the same for all 40 people - \frac{5}{40}, (initial probability of picking the pass (\frac{5}{40}) will be the same for any person in a row).

AbhayPrasanna wrote:
35 - lose, 5 - win

Pick 5 people to win => 40C5 = total number of outcomes.

Favorable outcome is : First pick the 40th person, then pick any other 4.
=> 1*40C4

So, probability = 40C5 / 40C4

= 40!*36!*4! / (35!*5!*40!)

= 36/(35*5)

= 36 / 175


This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is 1*C^4_{39}. Also probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes} and you wrote vise-versa.

So P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}.

Hope it helps.


You are finding the probability that the first 34 people will not win and the 35th person wins, which is clearly not what we were asked to get.
_________________

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 of picking last   [#permalink] 23 Dec 2012, 05:22
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic In a certain game, you pick a card from a standard deck of mikemcgarry 1 20 Dec 2012, 18:10
1 Getting from a 37 to the mid 40s futurectdoc 6 20 Jun 2012, 21:06
GMAT 550 any chances of getting picked in TOP 50 MBA univ vidhu 2 09 Nov 2011, 08:02
Pick a card from the deck-Probability soaringAlone 1 15 Jun 2011, 20:48
2 Probability Question - Picking the cards superglue 8 08 Aug 2009, 09:09
Display posts from previous: Sort by

If 40 people get the chance to pick a card from a canister

  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®.