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

It is currently 19 Oct 2018, 23:25

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

M23-30

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

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
M23-30  [#permalink]

Show Tags

New post 16 Sep 2014, 01:19
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (01:08) correct 37% (01:19) wrong based on 139 sessions

HideShow timer Statistics

A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?

A. \(\frac{1}{10}\)
B. \(\frac{1}{5}\)
C. \(\frac{3}{10}\)
D. \(\frac{2}{5}\)
E. \(\frac{1}{2}\)

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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?
Extra-hard Quant Tests with Brilliant Analytics

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
Re M23-30  [#permalink]

Show Tags

New post 16 Sep 2014, 01:19
Official Solution:

A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?

A. \(\frac{1}{10}\)
B. \(\frac{1}{5}\)
C. \(\frac{3}{10}\)
D. \(\frac{2}{5}\)
E. \(\frac{1}{2}\)


We need to find the probability of \(T\), \(NM\), \(NM\) (Tom, not Mary, not Mary).

\(P(T, NM, NM)=3*\frac{1}{6} * \frac{4}{5} * \frac{3}{4}=\frac{3}{10}\), we are multiplying by 3 since \(\{T, NM, NM\}\) scenario can occur in 3 different ways: \(\{T, NM, NM\}\), \(\{NM, T, NM\}\), or \(\{NM, NM, T\}\).


Answer: C
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 19 Nov 2013
Posts: 25
Location: India
Concentration: Strategy, Technology
WE: Information Technology (Computer Software)
Re: M23-30  [#permalink]

Show Tags

New post 22 Oct 2014, 21:19
I am getting ans as 1/2

we have to find the probability of getting (T,NM,NM)

NM,NM - 5C2=10

and total ways- 6C3=20

therefore - 10/20 =1/2
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
Re: M23-30  [#permalink]

Show Tags

New post 22 Oct 2014, 23:55
2
Director
Director
User avatar
Joined: 25 Apr 2012
Posts: 692
Location: India
GPA: 3.21
WE: Business Development (Other)
Premium Member Reviews Badge
Re: M23-30  [#permalink]

Show Tags

New post 24 Oct 2014, 01:01
1
1
Bunuel wrote:
A committee of 3 has to be formed randomly from a group of 6 people. If Tom and Mary are in this group of 6, what is the probability that Tom will be selected into the committee but Mary will not?

A. \(\frac{1}{10}\)
B. \(\frac{1}{5}\)
C. \(\frac{3}{10}\)
D. \(\frac{2}{5}\)
E. \(\frac{1}{2}\)


Committee of 3 can be formed out of 6 people in 6C3 ways or 20 ways

Now Consider committee in which Tom is already a member so out of remaining 4 members (excluding Mary)..2 can be selected in 4C2 ways = 6 ways..

So probability 6/20 or 3/10
_________________


“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2657
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Premium Member Reviews Badge
Re: M23-30  [#permalink]

Show Tags

New post 01 Jan 2015, 20:59
i solved it this way:
6C3 = 20
combinations to choose Tom = 3C1 = 3
combinations not Mary = 3C1 = 3

we add these two 3+3 = 6
6/20 = 3/10
Current Student
User avatar
Joined: 12 Aug 2015
Posts: 287
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)
M23-30  [#permalink]

Show Tags

New post 16 Jan 2016, 02:03
6!/3!3! = 20 total possible combinations. 5!/3!2! = 10 combinations when the 2 are chosen together. 4!/3!1! = 4 comb when the 2 are out.

hence 1 - 14/20 = 3/10 combinations when either is in the team.
_________________

KUDO me plenty

Intern
Intern
avatar
Joined: 05 Aug 2015
Posts: 42
Re: M23-30  [#permalink]

Show Tags

New post 13 Mar 2016, 01:10
The solution makes sense, but can someone please let me know what I did wrong?

Total number of ways = 6C3 = 6*5*4/(3*2) = 20 ways -- this represents order doesn't matter

Total number of ways to select Tom, Not Marie, Not Marie = 1*4*3/(3*2) = 2 ways -- divide by 3*2 b/c order doesn't matter (to match denominator)

Thus 2/20 = 1/10; why is this wrong??
_________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 6961
Re: M23-30  [#permalink]

Show Tags

New post 13 Mar 2016, 02:16
happyface101 wrote:
The solution makes sense, but can someone please let me know what I did wrong?

Total number of ways = 6C3 = 6*5*4/(3*2) = 20 ways -- this represents order doesn't matter

Total number of ways to select Tom, Not Marie, Not Marie = 1*4*3/(3*2) = 2 ways -- divide by 3*2 b/c order doesn't matter (to match denominator)

Thus 2/20 = 1/10; why is this wrong??


Hi,

you have gone wrong here..


in initial 6C3, 3*2 was there because 3 people were selected and these 3 can be arranged in 3! ways..
But in second case, you have already selected 1 out of 3, and you have to select 2 out of remaining 4..
these two can be arranged in 2! so you have to divide by 2! and not 3!=3*2..

second case is 6 people- 1 already selected and 1 left out ..


2 to be picked up from 4..
so 4C2=4!/2!2!=6..
ans 6/20=3/10
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Current Student
avatar
Joined: 29 Nov 2013
Posts: 5
Location: United States (CA)
GMAT 1: 690 Q49 V35
GMAT 2: 690 Q49 V35
GPA: 4
Re: M23-30  [#permalink]

Show Tags

New post 28 Jul 2016, 13:40
Hi Bunuel

Why are we not multiplying by 3! instead of 3 -- those two NM's (NM1, NM2) also can be chosen differently.. so there should be total of 6 ways to select them .

Please clarify.
Intern
Intern
User avatar
Joined: 11 Nov 2014
Posts: 36
Concentration: Marketing, Finance
WE: Programming (Computer Software)
M23-30  [#permalink]

Show Tags

New post 15 Aug 2016, 02:44
2
In my way,

I found this to be the easiest solution for the question.
ways of choosing 3 people out of 6 = 6C3
ways of choosing only Tom from Tom and Mary = 1
and choosing remaining 2 from remaining 4 (excluding tom and Mary) = 4C2

Probability of choosing 3 member including Tom but not Mary =
1 x 4C2
--------- = 3/10
6C3
Manager
Manager
avatar
S
Joined: 17 Aug 2015
Posts: 110
GMAT 1: 650 Q49 V29
Reviews Badge
Re: M23-30  [#permalink]

Show Tags

New post 07 Sep 2016, 13:34
this is not a permutation question, where is Tom position in the group of three doesnt matter

easiest way total combination 6c3

tom will occupy one spot and remaining 02 spots in the group will be filled from 04 person (05- not including mary)- 1*4c2

so probability 4c2/6c3
Intern
Intern
User avatar
B
Joined: 06 Feb 2016
Posts: 48
Location: Poland
Concentration: Finance, Accounting
GMAT 1: 730 Q49 V41
GPA: 3.5
Re: M23-30  [#permalink]

Show Tags

New post 20 Sep 2016, 06:54
I did it correctly and the difficulty level changed from hard to medium :x
Manager
Manager
avatar
B
Joined: 23 Nov 2016
Posts: 76
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51
Premium Member
Re: M23-30  [#permalink]

Show Tags

New post 22 Feb 2017, 11:58
mvictor wrote:
i solved it this way:
6C3 = 20
combinations to choose Tom = 3C1 = 3
combinations not Mary = 3C1 = 3

we add these two 3+3 = 6
6/20 = 3/10


This is incorrect. Consider the six people to be {A, B, C, D, Tom, Mary}

There are 10 combinations which include Tom, of which 6 do not include Mary: {A,B,Tom}; {B,C,Tom}; {C,D,Tom}; {A,C,Tom}; {A,D,Tom}; {B,D,Tom}; {A, Tom, Mary}; {B, Tom, Mary}; {C, Tom, Mary}; {D, Tom, Mary}.

Per the same logic above, there are 10 combinations which include Mary, which means there are 10 combinations that do not include Mary.

Combos with Tom only: 6
Combos with Mary only: 6
Combos with Tom and Mary: 4
Combos without Tom and Mary: 4
Manager
Manager
avatar
B
Joined: 23 Nov 2016
Posts: 76
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51
Premium Member
Re: M23-30  [#permalink]

Show Tags

New post 22 Feb 2017, 12:20
shasadou wrote:
6!/3!3! = 20 total possible combinations. 5!/3!2! = 10 combinations when the 2 are chosen together. 4!/3!1! = 4 comb when the 2 are out.

hence 1 - 14/20 = 3/10 combinations when either is in the team.


This isn't correct, there are only four combinations when the two are chosen together. For your approach, you'd have to do:

1-(Mary only)-(Tom and Mary)-(without Tom or Mary) = 1- (1c1*1c0*4c2)/6c3 - (2c2*4c1)/6c3 - (2c0*4c3)/6c3 = 6/20=3/10

Combos with Tom only: 6
Combos with Mary only: 6
Combos with Tom and Mary: 4
Combos without Tom and Mary: 4
Intern
Intern
avatar
B
Joined: 03 Jun 2017
Posts: 7
Re: M23-30  [#permalink]

Show Tags

New post 13 Jul 2017, 18:36
I've solved it this way:
Probability that Tom will be chosen is to fill up one of the three seats is 3/6, and probability that Mary will not be chosen is 3/5 (5 is the remaining seats available). When I multiply those two probabilities I get the following: 3/6 x 3/5 = 3/10
Answer C
Senior Manager
Senior Manager
User avatar
S
Joined: 08 Jun 2015
Posts: 451
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33
Premium Member
Re: M23-30  [#permalink]

Show Tags

New post 27 Mar 2018, 06:00
+1 for option C. The required probability is : 4c2/6c3.
_________________

" The few , the fearless "

Manager
Manager
avatar
B
Joined: 03 Oct 2016
Posts: 129
Re: M23-30  [#permalink]

Show Tags

New post 29 Mar 2018, 16:23
Group of People 6 ----->>>> Committee needs to be formed of 3 Members.

Mary is not required in the committee so keep her out. Group of people =6-1 (subtracting Mary from total) = 5.
Now we have to keep Tom. Probability of Tom for sure \(\frac{1}{5}\).
Mary is gone and now Tom has taken place in committee of 3 so, Group of people=5-1 (subtracting 1 for Tom as now he is committee)
Now the remaining 2 post of the committee can been filed in \(\frac{2}{4}\) ways.

Total probability = \(\frac{1}{5}\) * \(\frac{2}{4}\) = \(\frac{1}{10}\).

Wrong.

We need to multiple Total probability by 3 as Mr. Tom can be selected at any three places. _T_, T_ _ or _ _ T.

Hence, total probability = 3*\(\frac{1}{10}\) = \(\frac{3}{10}\). (C).
_________________

:-) Non-Allergic To Kudos :-)

GMAT Club Bot
Re: M23-30 &nbs [#permalink] 29 Mar 2018, 16:23
Display posts from previous: Sort by

M23-30

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

Moderators: chetan2u, Bunuel



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®.