It is currently 17 Nov 2018, 04:55
My Tests

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.

A club has exactly 3 men and 7 women as members. If two memb

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Moderator
Moderator
User avatar
Joined: 18 Apr 2015
Posts: 4895
Followers: 74

Kudos [?]: 976 [0], given: 4494

CAT Tests
A club has exactly 3 men and 7 women as members. If two memb [#permalink] New post 22 Jul 2018, 08:12
Expert's post
00:00

Question Stats:

75% (01:01) correct 25% (01:50) wrong based on 12 sessions
A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?

A. \(\frac{14}{15}\)

B. \(\frac{4}{5}\)

C. \(\frac{8}{15}\)

D. \(\frac{7}{15}\)

E. \(\frac{1}{5}\)
[Reveal] Spoiler: OA

_________________

Get the 2 FREE GREPrepclub Tests

1 KUDOS received
Director
Director
User avatar
Joined: 07 Jan 2018
Posts: 538
Followers: 4

Kudos [?]: 464 [1] , given: 82

CAT Tests
Re: A club has exactly 3 men and 7 women as members. If two memb [#permalink] New post 23 Jul 2018, 07:49
1
This post received
KUDOS
we can solve this problem by a direct method which is more time consuming.
i.e finding the no of ways that at least one women is selected.
However an easier approach is to find the no of ways in which only men are selected for both the position and reducing this probabilty by 1.

No. of ways in which only men can be selected for both the position is

\(3 * 2 = 6\) since there are \(3\) men and same person cannot hold both the position.
total no of possible outcome would be \(10 * 9 = 90\) This is because men + women = 10

therefore probability = \(\frac{6}{90} = \frac{2}{30} = \frac{1}{15}\)
Hence probability that at least a women is selected = \(\frac{15}{15} - \frac{1}{15} = \frac{14}{15}\)
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Re: A club has exactly 3 men and 7 women as members. If two memb   [#permalink] 23 Jul 2018, 07:49
Display posts from previous: Sort by

A club has exactly 3 men and 7 women as members. If two memb

  Question banks Downloads My Bookmarks Reviews Important topics  


GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.