It is currently 22 Sep 2017, 10:31

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

From a group of 8 volunteers, including Andrew and Karen

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

Hide Tags

12 KUDOS received
Manager
Manager
avatar
Joined: 14 Dec 2011
Posts: 218

Kudos [?]: 378 [12], given: 172

GPA: 3.46
WE: Information Technology (Consulting)
Premium Member
From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 02 Oct 2013, 06:28
12
This post received
KUDOS
56
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

71% (01:10) correct 29% (01:45) wrong based on 1010 sessions

HideShow timer Statistics

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

A. 3/7
B. 5/12
C. 27/70
D. 2/7
E. 9/35

[Reveal] Spoiler:
Here, I approached the problem as below:

Considering that Andrew will be there , we need to select 3 other people from 6 remaining volunteers excluding Karen.

=> 6C3 / 7C3 = 4/7.

Although finally I randomly guessed and selected correct answer i.e. 2/7, I was not able to get the answer with my approach mentioned here.

Could someone tell me what am I missing here?
[Reveal] Spoiler: OA

Kudos [?]: 378 [12], given: 172

12 KUDOS received
Intern
Intern
User avatar
Joined: 14 Aug 2013
Posts: 35

Kudos [?]: 91 [12], given: 4

Location: United States
Concentration: Finance, Strategy
GMAT Date: 10-31-2013
GPA: 3.2
WE: Consulting (Consumer Electronics)
GMAT ToolKit User
Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 02 Oct 2013, 07:28
12
This post received
KUDOS
3
This post was
BOOKMARKED
mba1382 wrote:
From a group of 8 volunteers,including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

A. 3/7
B. 5/12
C. 27/70
D. 2/7
E. 9/35

Here, I approached the problem as below:

Considering that Andrew will be there , we need to select 3 other people from 6 remaining volunteers excluding Karen.

=> 6C3 / 7C3 = 4/7.

Although finally I randomly guessed and selected correct answer i.e. 2/7, I was not able to get the answer with my approach mentioned here.

Could someone tell me what am I missing here?


=No.of ways in selecting 4 members group with Andrew and without Karein(6C3 Ways) / Total No of ways of selecting 4 member group from 8 volunteers (8C4 ways)
=>6C3/8C4
=>2/7
Ans D hope it helps

Kudos [?]: 91 [12], given: 4

Intern
Intern
avatar
Joined: 06 Feb 2013
Posts: 23

Kudos [?]: 6 [0], given: 3

Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 03 Oct 2013, 09:54
kusena wrote:
=No.of ways in selecting 4 members group with Andrew and without Karein(6C3 Ways) / Total No of ways of selecting 4 member group from 8 volunteers (8C4 ways)
=>6C3/8C4
=>2/7


Hi kusena,

could you please elaborate how you approach this kind of questions? what do 6C3 & 8C4 mean? :oops:
thanks

Kudos [?]: 6 [0], given: 3

9 KUDOS received
Manager
Manager
User avatar
Joined: 18 Dec 2012
Posts: 96

Kudos [?]: 54 [9], given: 34

Location: India
Concentration: General Management, Strategy
GMAT 1: 660 Q49 V32
GMAT 2: 530 Q37 V25
GPA: 3.32
WE: Manufacturing and Production (Manufacturing)
GMAT ToolKit User
Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 03 Oct 2013, 10:03
9
This post received
KUDOS
4
This post was
BOOKMARKED
LinaNY wrote:
kusena wrote:
=No.of ways in selecting 4 members group with Andrew and without Karein(6C3 Ways) / Total No of ways of selecting 4 member group from 8 volunteers (8C4 ways)
=>6C3/8C4
=>2/7


Hi kusena,

could you please elaborate how you approach this kind of questions? what do 6C3 & 8C4 mean? :oops:
thanks


Hi

The problem is a mix of combinations & probability.
8C4 means we can choose 4 people out of a group of 8 in 8C4 ways. 8C4 expands to 8x7x6x5/1x2x3x4
Similarly choosing a group of 4 with Andrew and no karein is 6C3 ways. Here the group is 6 because, we don't need Karein and Andrew is already in the group we want. Hence we need to choose 3 people from the remaining 6. Which is 6C3 ways.

Probability is 6C3 / 8C4 = (6x5x4/1x2x3)/(8x7x6x5/1x2x3x4) = 2/7 (answer)
_________________

I'm telling this because you don't get it. You think you get it which is not the same as actually getting it. Get it?

Kudos [?]: 54 [9], given: 34

Intern
Intern
avatar
Joined: 29 May 2012
Posts: 12

Kudos [?]: 114 [0], given: 18

Concentration: General Management, Finance
GPA: 3.49
Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 04 Oct 2013, 03:24
------ removed ------
_________________

"When ideas get really complicated and the world gets complicated, its foolish to think that the person who's first can figure it all out"


Last edited by accincognito on 31 Oct 2013, 05:38, edited 2 times in total.

Kudos [?]: 114 [0], given: 18

Expert Post
8 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41687

Kudos [?]: 124452 [8], given: 12078

Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 04 Oct 2013, 04:39
8
This post received
KUDOS
Expert's post
17
This post was
BOOKMARKED
paskorntt wrote:
Quote:
Hi

The problem is a mix of combinations & probability.
8C4 means we can choose 4 people out of a group of 8 in 8C4 ways. 8C4 expands to 8x7x6x5/1x2x3x4
Similarly choosing a group of 4 with Andrew and no karein is 6C3 ways. Here the group is 6 because, we don't need Karein and Andrew is already in the group we want. Hence we need to choose 3 people from the remaining 6. Which is 6C3 ways.

Probability is 6C3 / 8C4 = (6x5x4/1x2x3)/(8x7x6x5/1x2x3x4) = 2/7 (answer)


Still don't understand how you got the 6C3, please kindly explain.


From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?
A. 3/7
B. 5/12
C. 27/70
D. 2/7
E. 9/35

We need Andrew and 3 others but not Karen in the group. The # of ways to choose 3 members out of 8-2=6 (all but Andrew and Karen) is \(C^3_6\), thus \(P=\frac{C^3_6}{C^4_8}=\frac{2}{7}\).

Answer: D.
_________________

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

Kudos [?]: 124452 [8], given: 12078

21 KUDOS received
Intern
Intern
avatar
Joined: 08 Aug 2011
Posts: 22

Kudos [?]: 98 [21], given: 17

GMAT ToolKit User
Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 24 Oct 2013, 01:07
21
This post received
KUDOS
12
This post was
BOOKMARKED
You can also just find the probability of selecting the desired individuals and then multiply that by the number of ways those selections can be ordered:

P(Andrew)*P(Not Karen)*P(Not Karen)*P(Not Karen)*(ways to order selections) = \((\frac{1}{8})*(\frac{6}{7})*(\frac{5}{6})*(\frac{4}{5})*(\frac{4!}{3!}) = (\frac{1}{14})*4 = \frac{2}{7}\)

The above is just the sum of the individual probabilities of each possible 4-person team:

Andrew, Not Karen, Not Karen, Not Karen -->
\((\frac{1}{8})(\frac{6}{7})(\frac{5}{6})(\frac{4}{5}) = \frac{1}{14}\)

Not Karen or Andrew, Andrew, Not Karen, Not Karen -->
\((\frac{6}{8})(\frac{1}{7})(\frac{5}{6})(\frac{4}{5})= \frac{1}{14}\)

Not Karen or Andrew, Not Karen or Andrew, Andrew, Not Karen -->
\((\frac{6}{8})(\frac{5}{7})(\frac{1}{6})(\frac{4}{5})= \frac{1}{14}\)

Not Karen or Andrew, Not Karen or Andrew, Not Karen or Andrew, Andrew -->
\((\frac{6}{8})(\frac{5}{7})(\frac{4}{6})(\frac{1}{5})= \frac{1}{14}\)

Thus, \((\frac{1}{14})*4=\frac{2}{7}\)

Kudos [?]: 98 [21], given: 17

3 KUDOS received
Intern
Intern
avatar
Joined: 20 Oct 2012
Posts: 5

Kudos [?]: 7 [3], given: 2

Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 12 Nov 2013, 00:29
3
This post received
KUDOS
1
This post was
BOOKMARKED
Sorry, wanted to add this approach as an alternative.

Was helping someone with this question, and given that she suffers from over thinking a problem too much that it eats in to her exam time, she found the slot method easier to 'understand'... hope it helps those who are still having difficulties:


Find the # 4-man organizing parties from group of 8:

(8 x 7 x 6 x 5) / (4 x 3 x 2) = 70

- TOP: Four "slots". Number of members remaining to fill slot.
- BOTTOM: 4 slots to "shuffle".

Now find # of ways Andrew shows up without Karen:

(1) x (6 x 5 x 4) / (1) x (3 x 2) = 20

- TOP: "1" because only "1" Andrew. "6 x 5..." (1 less than above) because we don't want Karen.
- BOTTOM: 3 slots to "shuffle". Andrew represented by the "(1)"


Therefore, 20/70 = 2/7<------ Answer choice D.


I recommend looking for similar OG questions, and try using the above method a few times to get a feel for it.

Kudos [?]: 7 [3], given: 2

Intern
Intern
avatar
Joined: 28 Jan 2013
Posts: 34

Kudos [?]: 11 [0], given: 3

Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 26 Feb 2014, 16:59
uwengdori wrote:
From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random. What is the probability that Andrew will be among the 4 volunteers but not Karen?

3/7

5/12

27/70

2/7

9/35



out of the 4 members who are to be selected, there should be Andrew so 3 vacant positions remain.
Karen should not be selected, so no. of remaining persons 6(excluding Andrew and Karen)

3 people are to be selected for 3 Vacant positions from 6 people => 6C3 =>6!/(3!*3!) = 20

In general 4 people are to be selected from 8 people =>8C4 => 8!/(4!*4!)=70

Therefore the answer is 20/70= 2/7

Kudos [?]: 11 [0], given: 3

Intern
Intern
avatar
Joined: 06 Feb 2014
Posts: 3

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

Schools: Kellogg 1YR '15
Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 24 Jun 2014, 19:04
Can someone help me to solve this question?

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random. What is the probability that Andrew will be among the 4 selected AND Karen will not?

Answer is 2/7

I understand that there is 8!/4!4! ways to choose 4 volunteers from 8 people. I am struggling with how to then get the numerator, of the number of ways that Andrew can be chosen and Karen will not.

Thanks

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

Expert Post
11 KUDOS received
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7615

Kudos [?]: 16925 [11], given: 230

Location: Pune, India
Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 24 Jun 2014, 22:26
11
This post received
KUDOS
Expert's post
3
This post was
BOOKMARKED
achakrav2694 wrote:
Can someone help me to solve this question?

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random. What is the probability that Andrew will be among the 4 selected AND Karen will not?

Answer is 2/7

I understand that there is 8!/4!4! ways to choose 4 volunteers from 8 people. I am struggling with how to then get the numerator, of the number of ways that Andrew can be chosen and Karen will not.

Thanks



There are 8 people in the volunteers group. You need to choose 4.
You have already chosen Andrew. You kick out Karen. Now you are left with 6 people. To make a group of 4, you need 3 more people (Andrew is already in). From 6 people, how do you choose 3? In 6C3 (or 6!/3!*3!) = 20 ways.

So required probability = 20/(8!/4!4!) = 20/70 = 2/7
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 16925 [11], given: 230

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41687

Kudos [?]: 124452 [0], given: 12078

Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 25 Jun 2014, 02:36
achakrav2694 wrote:
Can someone help me to solve this question?

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random. What is the probability that Andrew will be among the 4 selected AND Karen will not?

Answer is 2/7

I understand that there is 8!/4!4! ways to choose 4 volunteers from 8 people. I am struggling with how to then get the numerator, of the number of ways that Andrew can be chosen and Karen will not.

Thanks


Merging similar topics. Please refer to the discussion above.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rules 1, 2, 3 and 8. Thank you.


_________________

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

Kudos [?]: 124452 [0], given: 12078

1 KUDOS received
Manager
Manager
User avatar
Joined: 20 Dec 2013
Posts: 131

Kudos [?]: 104 [1], given: 1

Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 01 Jan 2015, 09:14
1
This post received
KUDOS
Its Andrew (Other) (Other) (Other) = (1/8) (6/7) (5/6) (4/5) (4!/3!) = 2/7

(4!/3!) is the number of ways AOOO can arrange itself. Logically speaking Andrew can be chosen in four ways - first, second, third, fourth.
_________________

76000 Subscribers, 7 million minutes of learning delivered and 5.6 million video views

Perfect Scores
http://perfectscores.org
http://www.youtube.com/perfectscores

Kudos [?]: 104 [1], given: 1

Manager
Manager
User avatar
Status: Kitchener
Joined: 03 Oct 2013
Posts: 97

Kudos [?]: 26 [0], given: 144

Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE: Education (Education)
Premium Member
From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 10 Feb 2015, 08:17
mba1382 wrote:
From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

A. 3/7
B. 5/12
C. 27/70
D. 2/7
E. 9/35

[Reveal] Spoiler:
Here, I approached the problem as below:

Considering that Andrew will be there , we need to select 3 other people from 6 remaining volunteers excluding Karen.

=> 6C3 / 7C3 = 4/7.

Although finally I randomly guessed and selected correct answer i.e. 2/7, I was not able to get the answer with my approach mentioned here.

Could someone tell me what am I missing here?


Dear Bunuel, How I can answer this question by the way ALL- THE BAD

where all = 8C4 = 70

the bad is when Karen and Andrew both are among the 4 volunteers selected

2C2*2 =number of the ways to selected Karen and Andrew

6C2 =number of the ways to selected other volunteers

8C4- 2*2C2 * 6C2
________________ =

8C4

70 - 2*15
_________ = 40/70 = 4/7 I do not know where is my mistake here?
70
_________________

Click +1 Kudos if my post helped

Kudos [?]: 26 [0], given: 144

Intern
Intern
avatar
Joined: 31 Dec 2014
Posts: 10

Kudos [?]: 9 [0], given: 7

Location: United States
Concentration: Marketing
GMAT 1: 700 Q44 V41
Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 10 Feb 2015, 21:29
Andrew's chances of being selected: 4/8
If Andrew is selected, Karen's chances of not being selected: 4/7
(4/8) * (4/7) = 16/56 or 2/7

Kudos [?]: 9 [0], given: 7

Expert Post
SVP
SVP
User avatar
G
Joined: 08 Jul 2010
Posts: 1808

Kudos [?]: 2207 [0], given: 50

Location: India
GMAT: INSIGHT
WE: Education (Education)
Reviews Badge
Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 14 Nov 2015, 00:00
mba1382 wrote:
From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

A. 3/7
B. 5/12
C. 27/70
D. 2/7
E. 9/35



Let's say that there are 4 places to be occupied

Andrew can take one of 4 places in 4 ways
6 out of remaining 7 (leaving Karen) can take their places in 6, 5 and 4 ways respectively

so favorable cases = 4*6*5*4

Total ways to occupy the places = 8*7*6*5

Probability = 4*6*5*4 / 8*7*6*5 = 2/7

Answer: Option D
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Kudos [?]: 2207 [0], given: 50

Senior Manager
Senior Manager
User avatar
S
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 393

Kudos [?]: 459 [0], given: 53

Location: United States (CA)
Age: 37
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GRE 1: 337 Q168 V169
WE: Education (Education)
From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 07 May 2016, 22:23
Attached is a visual that should help. Quick tip: notice that in all 4 versions, the numerators and denominators are the same, so you only have to calculate the probability once, then multiply it by 4.
Attachments

Screen Shot 2016-05-08 at 11.29.16 AM.png
Screen Shot 2016-05-08 at 11.29.16 AM.png [ 125.66 KiB | Viewed 17041 times ]


_________________

Harvard grad and 770 GMAT scorer, offering high-quality private GMAT tutoring, both in-person and via Skype, since 2002.

McElroy Tutoring

Kudos [?]: 459 [0], given: 53

1 KUDOS received
Senior Manager
Senior Manager
User avatar
S
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 393

Kudos [?]: 459 [1], given: 53

Location: United States (CA)
Age: 37
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GRE 1: 337 Q168 V169
WE: Education (Education)
From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 08 May 2016, 11:38
1
This post received
KUDOS
courtesy of Reinfrank2011:

Andrew, Not Karen, Not Karen, Not Karen -->
\((\frac{1}{8})(\frac{6}{7})(\frac{5}{6})(\frac{4}{5}) = \frac{1}{14}\)

Not Karen or Andrew, Andrew, Not Karen, Not Karen -->
\((\frac{6}{8})(\frac{1}{7})(\frac{5}{6})(\frac{4}{5})= \frac{1}{14}\)

Not Karen or Andrew, Not Karen or Andrew, Andrew, Not Karen -->
\((\frac{6}{8})(\frac{5}{7})(\frac{1}{6})(\frac{4}{5})= \frac{1}{14}\)

Not Karen or Andrew, Not Karen or Andrew, Not Karen or Andrew, Andrew -->
\((\frac{6}{8})(\frac{5}{7})(\frac{4}{6})(\frac{1}{5})= \frac{1}{14}\)

Thus, \((\frac{1}{14})*4=\frac{2}{7}\)

Happy studies!
_________________

Harvard grad and 770 GMAT scorer, offering high-quality private GMAT tutoring, both in-person and via Skype, since 2002.

McElroy Tutoring

Kudos [?]: 459 [1], given: 53

Intern
Intern
avatar
Joined: 23 May 2016
Posts: 12

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

Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 01 Jun 2016, 17:42
Suppose the question had said to find the probability to pick both or pick none, how would we do that? Would it be:

Pick both: 6C2 / 8C4

Pick None: 6C4 / 8C4 ?

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

Intern
Intern
avatar
Joined: 01 May 2015
Posts: 46

Kudos [?]: 17 [0], given: 1

Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

Show Tags

New post 01 Jun 2016, 21:05
Total number of combinations possible = C(8,4)
Andrew is always there. So, this leaves us with 7 volunteers to choose from. But even among these 7 volunteers, Karen should not be chosen. So, this leaves us with 6 volunteers, from whom we have to choose 3 (because Andrew has anyway been chosen).

So, number of combinations = C(6,3)

Hence, probability = C(6,3)/C(8,4) = 2/7

Kudos [?]: 17 [0], given: 1

Re: From a group of 8 volunteers, including Andrew and Karen   [#permalink] 01 Jun 2016, 21:05

Go to page    1   2    Next  [ 32 posts ] 

    Similar topics Author Replies Last post
Similar
Topics:
33 EXPERTS_POSTS_IN_THIS_TOPIC Ben needs to form a committee of 3 from a group of 8 enginee eltonli 13 13 Jun 2017, 12:41
8 EXPERTS_POSTS_IN_THIS_TOPIC A teacher will pick a group of 4 students from a group of 8 amod243 12 06 Nov 2016, 10:03
8 EXPERTS_POSTS_IN_THIS_TOPIC From a group of 21 astronauts that includes 12 people with Stiv 5 04 Nov 2016, 09:07
4 EXPERTS_POSTS_IN_THIS_TOPIC From a group of 21 astronauts that includes 12 people with japped187 5 05 Sep 2013, 01:14
37 EXPERTS_POSTS_IN_THIS_TOPIC From a Group of 8 People, Including George and Nina, 3 people are mohitmohit11 15 29 Nov 2016, 16:10
Display posts from previous: Sort by

From a group of 8 volunteers, including Andrew and Karen

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


GMAT Club MBA Forum Home| About| Terms and Conditions| 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®.