GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 06 Jul 2020, 03:35

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

In a group of 10 girls, only 4 have blue eyes. If 3 girls ar

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

Hide Tags

Find Similar Topics 
Intern
Intern
avatar
Joined: 15 Jun 2010
Posts: 9
In a group of 10 girls, only 4 have blue eyes. If 3 girls ar  [#permalink]

Show Tags

New post Updated on: 16 Apr 2014, 23:52
1
3
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

65% (02:28) correct 35% (02:39) wrong based on 129 sessions

HideShow timer Statistics

In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5

Originally posted by tauchmeister on 16 Apr 2014, 14:55.
Last edited by Bunuel on 16 Apr 2014, 23:52, edited 1 time in total.
Edited the question
Manager
Manager
avatar
Status: suffer now and live forever as a champion!!!
Joined: 01 Sep 2013
Posts: 96
Location: India
Dheeraj: Madaraboina
GPA: 3.5
WE: Information Technology (Computer Software)
Re: In a group of 10 girls, only 4 have blue eyes.  [#permalink]

Show Tags

New post 16 Apr 2014, 18:17
atleast 2 girls means 2 girls (or) more than 2 girls.

Total = 10;
Having blue eyes =4;
3 girls are selected at random without replacement
probability that at least 2 girls will have blue eyes is 4c2*6c1(exacltly 2 girls)/10c3 + 4c3*6c0(more than 2 girls)/10c3
= 3/10 + 1/30 = 1/3;

Press kudos if this helps you :)
e-GMAT Representative
User avatar
V
Joined: 04 Jan 2015
Posts: 3410
In a group of 10 girls, only 4 have blue eyes. If 3 girls ar  [#permalink]

Show Tags

New post 20 Apr 2015, 02:06
2
1
tauchmeister wrote:
In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5


The Event here can happen in 2 ways:
i) 2 of the 3 girls have blue eyes and the 3rd girl has non-blue eyes
ii) All 3 of the 3 girls have blue eyes

Now, the important point that students must know in PnC and Probability questions is that OR scenarios imply ADDITION, and AND scenarios imply MULTIPLICATION.

In the current question, either Case (i) will happen OR Case (ii) will happen. The two cases cannot happen at the same time (You cannot have 2 girls out of the 3 girls having blue eyes AND 3 girls out of the 3 girls having blue eyes at the same time)

Since this is an OR scenario, we will ADD the probabilities of Case (i) and Case (ii).

So, we'll write: P(at least 2 blue-eyed girls) = P(2 blue-eyed girls) + P(3 blue-eyed girls) . . . (1)

Let's now find P(2 blue-eyed girls)

Here the favorable scenario is: 2 girls are blue-eyed AND 1 girl is non-blue-eyed

Since this is an AND scenario, the number of favorable cases will be obtained by multiplying the number of ways in which 2 blue-eyed girls can be selected (out of the 4 blue-eyed girls) and the number of ways in which 1 non-blue-eyed girl can be selected (out of the 6 non-blue-eyed girls)

So, P(2 blue-eyed girls) = (4C2) X(6C1)/(10C3) = 3/10 . . . (2)

P(3 blue-eyed girls) = (4C3)/(10C3) = 1/30. . . (3)

Subsituting (2) and (3) in (1), we get:

P(at least 2 blue-eyed girls) = \(\frac{3}{10}+\frac{1}{30}\) = 1/3

Hope this helped! :)

- Japinder
_________________
Current Student
avatar
S
Joined: 23 Jul 2015
Posts: 138
GMAT ToolKit User
Re: In a group of 10 girls, only 4 have blue eyes. If 3 girls ar  [#permalink]

Show Tags

New post 16 Jan 2017, 03:52
tauchmeister wrote:
In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5


Total selection = 10C3 = 10 * 3 * 4
Exactly 1 girl has blue eyes = 4C1 = 4
Exactly 3 girls don't have blue eyes = 6C3 = 5 * 4

At least 1 girl has blue eyes = \(1 - [(4 * 5 * 4)/ 10 * 3 * 4]\) = \(1 - 2/3\) =\(1/3\)
Ans.: B
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2799
Re: In a group of 10 girls, only 4 have blue eyes. If 3 girls ar  [#permalink]

Show Tags

New post 18 Jan 2017, 10:01
tauchmeister wrote:
In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5


We are given that we have a group of 10 girls; 4 girls have blue eyes and 6 girls do not have blue eyes. We need to determine the probability, when 3 girls are selected, that at least 2 girls will have blue eyes. There are two scenarios in which we can select at least 2 girls with blue eyes.

Scenario 1: Blue-Blue-Not Blue

Scenario 2: Blue-Blue-Blue

Before calculating each scenario, let’s first determine the total number of ways to select 3 girls from a group of 10 girls.

The number of ways to select 3 girls from a group of 10 is:

10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2 x 1) = 5 x 3 x 8 = 120

Next we need to determine the number of ways to select the blue-eyed girls.

First, let’s calculate scenario 1: Blue-Blue-Not Blue

The number of ways to select 2 girls with blue eyes from 4 is:

4C2 = 4!/2!2! = (4 x 3)/2! = 6 ways

The number of ways to select 1 girl with no blue eyes from 6 is:

6C1 = 6 ways

Thus, the probability of selecting 2 girls with blue eyes and 1 with no blue eyes is:

(6 x 6)/120 = 36/120

Next, let’s calculate scenario 2: Blue-Blue-Blue

The number of ways to select 3 girls with blue eyes from 4 is:

4C3 = (4 x 3 x 2)/3! = 4 ways

Thus, the probability of selecting 3 girls with blue eyes is:

4/120

Finally, we can determine the probability of selecting at least 2 girls with blue eyes:

36/120 + 4/120 = 40/120 = 1/3

Answer: B
_________________

Jeffrey Miller

Head of GMAT Instruction

Jeff@TargetTestPrep.com
225 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

GMAT Club Bot
Re: In a group of 10 girls, only 4 have blue eyes. If 3 girls ar   [#permalink] 18 Jan 2017, 10:01

In a group of 10 girls, only 4 have blue eyes. If 3 girls ar

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





cron

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