Last visit was: 23 Jul 2024, 10:31 It is currently 23 Jul 2024, 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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94592
Own Kudos [?]: 643301 [11]
Given Kudos: 86728
Send PM
CEO
CEO
Joined: 26 Feb 2016
Posts: 2863
Own Kudos [?]: 5332 [3]
Given Kudos: 47
Location: India
GPA: 3.12
Send PM
Intern
Intern
Joined: 09 Apr 2017
Posts: 2
Own Kudos [?]: 3 [0]
Given Kudos: 33
Send PM
Target Test Prep Representative
Joined: 04 Mar 2011
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Posts: 3036
Own Kudos [?]: 6619 [2]
Given Kudos: 1646
Send PM
Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]
1
Kudos
1
Bookmarks
Expert Reply
Bunuel wrote:
From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected?

(A) 0.1

(B) 0.2

(C) 0.25

(D) 0.4

(E) 0.6


There are a total of 5C2 = 5!/(3! x 2!) = (5 x 4)/2 = 10 ways of choosing two people from a group of five people. Since the selection of Marnie and Noomi corresponds to one of these choices, the probability of this selection is 1/10 = 0.1.

Answer: A
GMAT Club Legend
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30853 [4]
Given Kudos: 799
Location: Canada
Send PM
Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]
2
Kudos
2
Bookmarks
Expert Reply
Top Contributor
Bunuel wrote:
From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected?

(A) 0.1

(B) 0.2

(C) 0.25

(D) 0.4

(E) 0.6


Our goal is to find P(M and N both selected)

There are two ways to approach this.

Method #1:
P(M and N both selected) = P(one of them is selected 1st AND the other selected 2nd)
= P(one of them is selected 1st) x P(the other selected 2nd)
= (2/5)(1/4)
= 1/10
= 0.1
Aside: P(one of them is selected 1st) = 2/5 because I'm allowing for either Marnie or Noomi to be selected first.

Method #2:
P(M and N both selected) = P(M selected 1st AND N selected 2nd OR N selected 1st AND M selected 2nd)
= P(M selected 1st AND N selected 2nd) + P(N selected 1st AND M selected 2nd)
= (1/5)(1/4) + (1/5)(1/4)
= 1/20 + 1/20
= 1/10
= 0.1

Cheers,
Brent
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 34045
Own Kudos [?]: 853 [0]
Given Kudos: 0
Send PM
Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]
Moderator:
Math Expert
94592 posts