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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
02 Sep 2013, 01:14

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

70% (02:46) correct
30% (01:46) wrong based on 147 sessions

Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

Re: Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
02 Sep 2013, 09:38

4

This post received KUDOS

Expert's post

arakban99 wrote:

Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

(A) 1/3 (B) 2/5 (C) 1/2 (D) 3/5 (E) 2/3

Probability = \(\frac{Desired Outcomes}{Total Outcomes}\)

Event = Selection of 3 employees to assign office from 5 such that 2 are men and 1 is woman

Total Outcomes = Selection of any 3 employees from 5 employees = 5 C 3 = 10

Desired Outcomes = Selection of 2 Men from 3 Men AND selection of 1 woman from 2 Women. = (3 C 2) * (2 C 1) = 3*2 = 6

Probability = \(\frac{6}{10}\) --------> \(\frac{3}{5}\) _________________

Re: Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
06 Sep 2013, 22:06

i guessed that it at least more than 1/2, eliminated ABC and chose D. Combination approach looks best because i got crazy to do probability one. I used reverse one, which is 1 - (3/5*2/4*1/3+1/5*1/4*1+3/5*1/4*1/3+2/5*3/4*1/3)= 1- (1/10+1/10+1/10+1/10)=6/10=3/5

Re: Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
07 Sep 2013, 04:47

1

This post received KUDOS

Temurkhon wrote:

i guessed that it at least more than 1/2, eliminated ABC and chose D. Combination approach looks best because i got crazy to do probability one. I used reverse one, which is 1 - (3/5*2/4*1/3+1/5*1/4*1+3/5*1/4*1/3+2/5*3/4*1/3)= 1- (1/10+1/10+1/10+1/10)=6/10=3/5

write if it is right

Well, your method is good, but I think this is a more safe method:

The committee should look like this : Man, Man, Woman

how to choose 3 people to be assigned in office from 5: 5C3 = 5!/3!2! = 10

How to choose 2 man and 1 woman= 3C2 * 2C1 = 3!/2! * 2!/1! = 6

Probability = Number of Desired outcome /Total possible outcome = 6/10 = 3/5

Re: Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
08 Sep 2013, 01:14

1

This post received KUDOS

offices can be allocated as m,m,w or m,w,m or w,m,m P(m,m,w) = 3/5*2/4*2/3 P(m,w,m) = 3/5*2/4*2/3 P(w,m,m) = 2/5*3/4*2/4

Since its 'OR', need to add all the 3 probabilities. So answer is 1/5+1/5+1/5 = 3/5 somehow am not comfortable with combinations, hence chose to take a direct method. _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
06 Nov 2013, 15:47

Narenn wrote:

arakban99 wrote:

Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

(A) 1/3 (B) 2/5 (C) 1/2 (D) 3/5 (E) 2/3

Probability = \(\frac{Desired Outcomes}{Total Outcomes}\)

Event = Selection of 3 employees to assign office from 5 such that 2 are men and 1 is woman

Total Outcomes = Selection of any 3 employees from 5 employees = 5 C 3 = 10

Desired Outcomes = Selection of 2 Men from 3 Men AND selection of 1 woman from 2 Women. = (3 C 2) * (2 C 1) = 3*2 = 6

Probability = \(\frac{6}{10}\) --------> \(\frac{3}{5}\)

Hi Narenn, can you explain why the total outcomes is not 5! , (the number of ways to arrange 5 different objects)?

Re: Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
21 Nov 2013, 00:45

1

This post received KUDOS

oss198 wrote:

Narenn wrote:

arakban99 wrote:

Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

(A) 1/3 (B) 2/5 (C) 1/2 (D) 3/5 (E) 2/3

Probability = \(\frac{Desired Outcomes}{Total Outcomes}\)

Event = Selection of 3 employees to assign office from 5 such that 2 are men and 1 is woman

Total Outcomes = Selection of any 3 employees from 5 employees = 5 C 3 = 10

Desired Outcomes = Selection of 2 Men from 3 Men AND selection of 1 woman from 2 Women. = (3 C 2) * (2 C 1) = 3*2 = 6

Probability = \(\frac{6}{10}\) --------> \(\frac{3}{5}\)

Hi Narenn, can you explain why the total outcomes is not 5! , (the number of ways to arrange 5 different objects)?

the question is asking for the probability of SELECTING 2 men & 1 woman, it does not matter if the selection includes woman A OR B. hope it helps. _________________

Re: Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
21 Nov 2013, 06:47

1

This post received KUDOS

Expert's post

oss198 wrote:

Hi Narenn, can you explain why the total outcomes is not 5! , (the number of ways to arrange 5 different objects)?

Hi oss198, Sorry, I saw your post today only.

Arrangement is out of context here. We are finding the ways of selecting the 3 persons out of 5. (either you choose ABC or BCA from ABCDE is same here)

Selection of 3 persons from 5 :- 5 C 3 selection of 3 persons from 5 and arranging them on 3 chairs :- 5 C 3 * 3! Selection of 5 persons from 5 :- 5 C 5 Selection of 5 persons from 5 and arranging them on 5 chairs :- 5 C 5 * 5! --------> which ultimately turns out to be 5!

If you need brief insights of these concepts check my article, Permutations and Combinations (In my Signature).

Re: Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
12 Dec 2013, 23:01

1

This post received KUDOS

1) 3/5*2/4*2/3=2/10 (where 3/5 probability of 1 man being selected, 2/4 the other man, 2/3 one woman) 2) 2/10*3=3/5 (multiplying by 3 as there are 3 cases in different order)

gmatclubot

Re: Of 5 employees, 3 are to be assigned an office and 2 are to
[#permalink]
12 Dec 2013, 23:01

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...