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

It is currently 25 Sep 2018, 18:04

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

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

Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2835
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 15 Sep 2016, 08:05
1
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


The probability that we’re looking for can be expressed as a follows:

# ways the group can be selected with Andrew but not Karen/total # of ways to select the group

Let’s begin with the numerator. If Andrew makes the team but Karen does not, we can mentally remove Andrew and Karen from the pool of people available for the group. This leaves 6 people available for the full 3 spots (remember Andrew must be selected). 3 people can be chosen from a group of 6 people in 6C3 ways:

(6 x 5 x 4)/3! = 20 ways

For the denominator, we’re looking for the total number of ways in which the 4-person team can be made. The team can be made in 8C4 ways or:

(8 x 7 x 6 x 5)/4! = 70 ways

Thus, the probability that Andrew is selected for the group but Karen is not is 20/70 = 2/7.

Answer: D
_________________

Jeffery Miller
Head of GMAT Instruction

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

Current Student
avatar
Joined: 29 Apr 2014
Posts: 11
Concentration: Entrepreneurship, Sustainability
GMAT 1: 710 Q46 V41
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 03 Oct 2016, 00:53
Bunuel

Definitely not my strong area but need to understand conceptually why this is wrong.

for andrew to be included - 8C1
For the rest three positions left 6C3

why is the 8C1 not required?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49496
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 03 Oct 2016, 02:17
goenkashreya wrote:
Bunuel

Definitely not my strong area but need to understand conceptually why this is wrong.

for andrew to be included - 8C1
For the rest three positions left 6C3

why is the 8C1 not required?


There are not 8 ways to choose Andrew. The probability of choosing Andrew out of 8 people is 1/8 but the number of ways to choose Andrew is 1 (1 out of 1 Andrew)..
_________________

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
B
Joined: 30 Apr 2015
Posts: 2
GPA: 3.92
WE: Information Technology (Other)
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 29 Oct 2016, 12:21
For me solving with probability using 2 steps is the easiest:

P(Andrew in 1 slot) = \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}\)

Andrew OR Andrew OR Andrew OR Andrew

P(Andrew in 1 slot & Karen in none) = \(\frac{1}{2} * \frac{6}{7} * \frac{5}{6} * \frac{4}{5} = \frac{2}{7}\)

Andrew AND not Karen AND not Karen AND not Karen

Is there a list of similar questions available other than searching for the "Probability" tag?
Intern
Intern
avatar
Joined: 27 Mar 2016
Posts: 1
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 30 Oct 2016, 06:28
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


Ans:-

A: Number of ways to select any 4 volunteers at random= 8C4
B: Number of Ways to Select Andrew= 1C1

Now that we have selected 1 volunteer, only 3 more volunteers are to be selected from a lot of 6 volunteers. [i.e. 8-1(Andrew)-1(Karen)=6volunteers]
C: Number of ways to select 4 Volunteers including Andrew but not Karen= 1C1*6C3= 6C3

Final Ans.: 6C3/8C4= 2/7 i.e. Option D
Manager
Manager
User avatar
G
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 65
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 31 Oct 2016, 02:27
Here is the simplet way to do it.
Attachments

thumbnail_IMAG0314.jpg
thumbnail_IMAG0314.jpg [ 87.1 KiB | Viewed 1279 times ]


_________________

Cours particuliers de GMAT

Intern
Intern
avatar
B
Joined: 04 Jan 2017
Posts: 7
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 13 Jan 2017, 10:47
[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
quote][/quote]

I solved this question in exactly the same way as decadecaf did (quoted above, sorry if it went wrong, first time quoting, new to GMAT club).
Is this method incorrect and simply a result of luck that I achieved the correct answer using this method?

Theory states that NOT questions and AT LEAST questions can be solved using complementary probability, I just want to know whether or not it is vital to use the combination methods to achieve the correct answer in this question.

Thanks for feedback in advance.
Manager
Manager
avatar
B
Joined: 23 Dec 2013
Posts: 179
Location: United States (CA)
GMAT 1: 710 Q45 V41
GMAT 2: 760 Q49 V44
GPA: 3.76
Reviews Badge
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 22 Jul 2017, 18:32
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?


In this case, we can start by finding the total number of choices: 8C4 = 8*7*6*5*4*3*2/4*3*2*4*3*2 = 14*5 = 70 choices.

The number of choices of a group with Andrew and without Karen is 6C3 because Andrew is already in the group (so we're down to 7 choices) and Karen is not in the group at all (so we're now down to 6 volunteers). We also deduct one from the choice (from 4 to 3) because Andrew has already been selected. 6C3 = 6*5*4*3*2/3*2*3*2 = 20.

20/70 = 2/7
Intern
Intern
avatar
B
Joined: 29 Nov 2016
Posts: 14
Location: United States (TX)
GMAT 1: 690 Q47 V38
GPA: 3.58
From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 21 Aug 2017, 18:25
Probability of selecting Andrew = 4 / 8
Then, probability of NOT selecting Karen = 4 / 7

(4 / 8) * (4 / 7 )
==> 4 / 14
==> 2 / 7

Am I missing something?
Manager
Manager
avatar
B
Joined: 01 Nov 2016
Posts: 68
Concentration: Technology, Operations
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 05 Oct 2017, 17:56
dadoprso wrote:
Probability of selecting Andrew = 4 / 8
Then, probability of NOT selecting Karen = 4 / 7

(4 / 8) * (4 / 7 )
==> 4 / 14
==> 2 / 7

Am I missing something?


Yes, the probability of selecting Andrew is not 4/8. There are 8 people and Andrew is one person. So probability of selecting Andrew is 1/8.
Intern
Intern
avatar
B
Joined: 08 Dec 2017
Posts: 17
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 09 Dec 2017, 09:12
Think of the selection as slots from left to right. We want andrew, not karen, not karen, not karen.

(1/8)(6/7)(5/6)(4/5)=
(1/8)(4/7)
We have to multiply by four to account for permutations. (1/8)(4/7)(4)=2/7

Posted from my mobile device
Senior Manager
Senior Manager
User avatar
S
Joined: 05 Dec 2016
Posts: 251
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
GMAT ToolKit User
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 14 Dec 2017, 01:30
total: 8C4=70
andrew within the selected group and Karen is not: 1*6C3=20
20/70=2/7
Answer is D
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12459
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 30 Jan 2018, 14:12
Hi All,

There's another way to "do the math" behind this question, although it takes a little longer - we can treat this question as a straight probability question and keep track of ALL the possible ways to get "what we want"….

We have 8 people to choose from; we'll pick 4 - we WANT Andrew in our group, we DON'T WANT Karen in our group.

_ _ _ _

If we want Andrew to be the first person we pick, then the other 3 CAN'T be Karen. We end up with the following math (notice that after picking each person, the total number of people remaining decreases):

(1/8)(6/7)(5/6)(4/5) = 4/56 = 1/14

Andrew COULD be the second, third or fourth person chosen though. Here's the math if he's SECOND and we don't pick Karen...

(6/8)(1/7)(5/6)(4/5) = 4/56 = 1/14

Notice how the product is the SAME.

If he's third, it would still be 1/14
if he's fourth, it would still be 1/14

With ALL of those possibilities, we end up with 1/14 + 1/14 +1/14 + 1/14 = 4/14 = 2/7

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Intern
Intern
User avatar
B
Joined: 27 Feb 2018
Posts: 8
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 28 Mar 2018, 05:24
jcaine wrote:
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.


I usually employ this method for all combination/permuation problems. Way easier!

Anyone can clarify why, when finding the number of ways Andrew shows up without Karen, we divide by 3! (3 x 2) instead of 4! (4 x 3 x 2) ?

We are shuffling all 4 slots. If order did matter Andrew - Member A - Member B - Member C would have been a different arrangement than Member A - Andrew - Member B - Member C

Thanks
Director
Director
User avatar
P
Joined: 09 Mar 2016
Posts: 886
From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post Updated on: 01 Jun 2018, 03:16
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?


Hey pushpitkc

can you please advise if my approach is correct ?

1/8*6/7*5/6*4/5=1/14 ---> 1/14*4 = 2/7 ( i multiply by 4 because there are 4 numbers of ways for andrew to be in the team ) i mean Andrew can be first, second, third or fourth

is my reasoning correct ?

what would be other way to solve through probability ?

have a great night :)
_________________

In English I speak with a dictionary, and with people I am shy.


Originally posted by dave13 on 31 May 2018, 14:24.
Last edited by dave13 on 01 Jun 2018, 03:16, edited 1 time in total.
BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3141
Location: India
GPA: 3.12
Premium Member CAT Tests
Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

Show Tags

New post 31 May 2018, 14:55
1
dave13 wrote:
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?


Hey pushpitkc

can you please advise if my approach is correct ?

1/8*6/7*5/6*4/5=1/14 ---> 1/4*4 = 2/7 ( i multiply by 4 because there are 4 numbers of ways for andrew to be in the team ) i mean Andrew can be first, second, third or fourth

is my reasoning correct ?

what would be other way to solve through probability ?

have a great night :)


Hey dave13

Your approach is perfect here and so is your reasoning. You have made a small typo!
_________________

You've got what it takes, but it will take everything you've got

GMAT Club Bot
Re: From a group of 8 volunteers, including Andrew and Karen &nbs [#permalink] 31 May 2018, 14:55

Go to page   Previous    1   2   [ 36 posts ] 

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  

Events & Promotions

PREV
NEXT


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®.