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

It is currently 14 Oct 2019, 20:57

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

According to study in an office, 12 out of 30 employees were intereste

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

Hide Tags

Find Similar Topics 
Director
Director
avatar
V
Joined: 18 Feb 2019
Posts: 580
Location: India
GMAT 1: 460 Q42 V13
GPA: 3.6
According to study in an office, 12 out of 30 employees were intereste  [#permalink]

Show Tags

New post 05 Mar 2019, 01:35
4
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

49% (02:37) correct 51% (03:01) wrong based on 88 sessions

HideShow timer Statistics

According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home?

A. 11/203
B. 192/203
D. 64/65
D. 1/203
E. None of these
Intern
Intern
avatar
B
Joined: 26 Dec 2018
Posts: 21
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

Show Tags

New post 05 Mar 2019, 22:28
I cannot clearly understand how the solution has been arrived at. Someone please explain

Posted from my mobile device
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58323
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

Show Tags

New post 05 Mar 2019, 23:21
2
2
kiran120680 wrote:
According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home?

A. 11/203
B. 192/203
D. 64/65
D. 1/203
E. None of these


We are selecting 3 employees and want at most 2 of them to be interested in doing work from home. At most 2 means 0, 1 or 2. So, any number but 3. Let's calculate the opposite probability and subtract that from 1.

APPROACH 1:

\(P(at \ most \ 2) = 1 - P(3) = 1 - \frac{12}{30}*\frac{11}{29}*\frac{10}{28} = \frac{192}{203}\).

Answer: B.

APPROACH 2:

\(P(at \ most \ 2) = 1 - P(3) = 1 - \frac{12C3}{30C3} = \frac{192}{203}\).

Answer: B.
_________________
GMAT Club Legend
GMAT Club Legend
User avatar
D
Joined: 18 Aug 2017
Posts: 4987
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
GMAT ToolKit User Premium Member
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

Show Tags

New post 06 Mar 2019, 01:36
Bunuel
quoting your line
\(P(at \ most \ 2) = 1 - P(3)

why have you subtracted the P of choosing employees who are interested in doing work from home by 1 ; wont that give us P of employees who do not want to work from home?

P of employee who want to work from home = 12/30
and P of employee who do not to work from home = 18/30
so
the probability that at most 2 of them were interested in doing work from home should be = 18/30 * 12/30 * 11/29

bunuel why is the highlighted part incorrect?

Bunuel wrote:
kiran120680 wrote:
According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home?

A. 11/203
B. 192/203
D. 64/65
D. 1/203
E. None of these


We are selecting 3 employees and want at most 2 of them to be interested in doing work from home. At most 2 means 0, 1 or 2. So, any number but 3. Let's calculate the opposite probability and subtract that from 1.

APPROACH 1:

[m]P(at \ most \ 2) = 1 - P(3) = 1 - \frac{12}{30}*\frac{11}{29}*\frac{10}{28} = \frac{192}{203}\).

Answer: B.

APPROACH 2:

\(P(at \ most \ 2) = 1 - P(3) = 1 - \frac{12C3}{30C3} = \frac{192}{203}\).

Answer: B.

_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Manager
Manager
avatar
S
Joined: 17 May 2018
Posts: 140
Location: India
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

Show Tags

New post 06 Mar 2019, 11:13
The method I did in is much longer but might be helpful
Atmost 2 means selecting 0, 1 , 2.
So case 1: we select 2 employees interested in work from home(wfh) and 1 employee mot interested in wfh
12c2*18c1
Case 2: we select 1 employee interested in work from home(wfh) and 2 employees mot interested in wfh
12c1*18c2
Case 3: we select 0 employees interested in work from home(wfh) and 3 employees mot interested in wfh
12c0*18c3
Therefore probability of atmost 2 people who want wfh being selected in 3 is sum of case 1,2 and 3, divided by total possibility of choosing any 3 people.
(12c2*18c1+12c1*18c2+12c0*18c3)/30c3
= 192/203
IMO B

Posted from my mobile device
Intern
Intern
avatar
B
Joined: 11 May 2018
Posts: 24
Location: India
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

Show Tags

New post 07 Mar 2019, 09:41
AjjayKannan wrote:
I cannot clearly understand how the solution has been arrived at. Someone please explain

Posted from my mobile device



question asks if we select 3 out of 30, what is probablity that at most 2 of them were interested in WfH
we can answer this by finding probablity of all of the selected are interested in WfH and subtracting it from 1. --> P(A) = 1 - P(A')

interested =12, not interested = 18
Number of ways to select 3 from 30 = 30C3
Number of ways to select 3 from 12 interested = 12C3

Probablity of all 3 selected are interested = 12C3/30C3

Probablity of at most two of the selected are interested = 1 - 12C3/30C3 = 192/203
Target Test Prep Representative
User avatar
D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8043
Location: United States (CA)
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

Show Tags

New post 10 Mar 2019, 19:50
kiran120680 wrote:
According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home?

A. 11/203
B. 192/203
D. 64/65
D. 1/203
E. None of these


Note that “at most 2” means “2 or fewer.” We can use the following formula:

P(at most 2 of them were interested in working from home) = 1 - P(all 3 of them were interested in working from home)

We have:

P(all 3 of them were interested in working from home) = 12/30 x 11/29 x 10/28 = 3/3 x 11/29 x 1/7 = 11/203

Therefore,

P(at most 2 of them were interested in working from home) = 1 - 11/203 = 203/203 - 11/203 = 192/203

Answer: B
_________________

Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Manager
Manager
avatar
G
Joined: 19 Sep 2017
Posts: 204
Location: United Kingdom
GPA: 3.9
WE: Account Management (Other)
GMAT ToolKit User Premium Member
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

Show Tags

New post 01 Apr 2019, 07:20
kiran120680 wrote:
According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home?

A. 11/203
B. 192/203
D. 64/65
D. 1/203
E. None of these

Hi,
Complimentary probability could solve this problem in the simplest possible way.

at most 2 will be in YYN,YNY,NYY cases. Y: yes, want to work from home, N: otherwise.
This shows that there will be ATLEAST one N.
We know that to find atleast = 1-None
None here will be YYY
So all YYY = 12/30 * 11/29 * 10/28
Atleast one N = atmost 2 Y = 1-(12/30*11/29*10/28)
= 1- 11/203 = 192/203

B
_________________
Cheers!!
GMAT Club Bot
Re: According to study in an office, 12 out of 30 employees were intereste   [#permalink] 01 Apr 2019, 07:20
Display posts from previous: Sort by

According to study in an office, 12 out of 30 employees were intereste

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





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne