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

It is currently 18 Oct 2018, 22:13

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

A Jar of marble contains 6 yellow and 4 blue marbles, and no other mar

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

Hide Tags

Manager
Manager
User avatar
B
Joined: 16 Mar 2016
Posts: 129
Location: France
GMAT 1: 660 Q47 V33
GPA: 3.25
GMAT ToolKit User
A Jar of marble contains 6 yellow and 4 blue marbles, and no other mar  [#permalink]

Show Tags

New post 03 May 2016, 13:43
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

69% (02:34) correct 31% (03:03) wrong based on 88 sessions

HideShow timer Statistics

Hello everyone,

I try to solve a probability question trough 2 different approaches : proba, AND combinatorics ...
I don't understand how to do it with combinatorics . Can you help me please ? Or do you have another and more efficient way ? :)

This is the problem :

A Jar of marble contains 6 yellow and 4 blue marbles, and no other marbles. If four are pulled out of the jar, one after the other, without being replaced, what is the probability that exactly two of the four marbles will be yellow and two blue.


A 1/10
B 1/7
C 1/6
D 3/7
9 9/11


My answer number :

There is 6 possibilities to arrange 2 yellow and 2 blue (let's write YYBB) according to the MISSISSIPPI rule : \(\frac{4 !}{2 x 2}\) = 6

Here are the 6 different possibilities :
YYBB, YBYB, YBBY, BYBY, BYYB, BBYY

So now, let's calculate the probabilities :

P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = P (YYBB) + P(YBYB) + P(YBBY + P(BYBY) + P(BYYB) + P(BBYY ), since the outcomes are all mutually exclusives.

P(YYBB) =\(\frac{6 * 5 * 4 * 3}{10 * 9 * 8 * 7}\) = \(\frac{1}{14}\)

Each expression is equal so : P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = 6* \(\frac{1}{14}\) =\(\frac{3}{7}\)
Manager
Manager
avatar
Joined: 12 Jun 2015
Posts: 79
Reviews Badge
Re: A Jar of marble contains 6 yellow and 4 blue marbles, and no other mar  [#permalink]

Show Tags

New post 03 May 2016, 13:59
3
4 marbles out of the given 10 are pulled out of the jar, one after the other, without being replaced

The number of possible combinations (total events) = 10C4

We need to find the probability that exactly two of the four marbles will be yellow(out of 6) and two blue( out of 4).

Here the order in which the marbles are removed isn't asked. So no permutation reqd.

The number of possible combinations to such a specific event = 6C2 X 4C2

The Probability of the specific event = [ 6C2 X 4C2 ] / 10C4

= 15 X 6 / 210
= 3/7
Correct Option : D
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50000
Re: A Jar of marble contains 6 yellow and 4 blue marbles, and no other mar  [#permalink]

Show Tags

New post 03 May 2016, 14:01
Alex75PAris wrote:
Hello everyone,

I try to solve a probability question trough 2 different approaches : proba, AND combinatorics ...
I don't understand how to do it with combinatorics . Can you help me please ? Or do you have another and more efficient way ? :)

This is the problem :

A Jar of marble contains 6 yellow and 4 blue marbles, and no other marbles. If four are pulled out of the jar, one after the other, without being replaced, what is the probability that exactly two of the four marbles will be yellow and two blue.


A 1/10
B 1/7
C 1/6
D 3/7
9 9/11


My answer number :

There is 6 possibilities to arrange 2 yellow and 2 blue (let's write YYBB) according to the MISSISSIPPI rule : \(\frac{4 !}{2 x 2}\) = 6

Here are the 6 different possibilities :
YYBB, YBYB, YBBY, BYBY, BYYB, BBYY

So now, let's calculate the probabilities :

P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = P (YYBB) + P(YBYB) + P(YBBY + P(BYBY) + P(BYYB) + P(BBYY ), since the outcomes are all mutually exclusives.

P(YYBB) =\(\frac{6 * 5 * 4 * 3}{10 * 9 * 8 * 7}\) = \(\frac{1}{14}\)

Each expression is equal so : P (YYBB or YBYB or YBBY or BYBY or BYYB or BBYY ) = 6* \(\frac{1}{14}\) =\(\frac{3}{7}\)


PROBABILITY:

\(\frac{4!}{2!*2!}*\frac{4}{10}*\frac{3}{9}*\frac{6}{8}*\frac{5}{7} =\frac{3}{7}\).

COMBINATORICS:

\(\frac{C^2_4*C^2_6}{C^4_{10}}=\frac{3}{7}\).

Answer: D.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
User avatar
B
Joined: 16 Mar 2016
Posts: 129
Location: France
GMAT 1: 660 Q47 V33
GPA: 3.25
GMAT ToolKit User
Re: A Jar of marble contains 6 yellow and 4 blue marbles, and no other mar  [#permalink]

Show Tags

New post 03 May 2016, 14:07
thank you ! What is the level of this question ?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50000
Re: A Jar of marble contains 6 yellow and 4 blue marbles, and no other mar  [#permalink]

Show Tags

New post 03 May 2016, 14:10
Intern
Intern
avatar
B
Joined: 05 Mar 2018
Posts: 32
CAT Tests
Re: A jar of marbles contains 6 yellow and 4 blue  [#permalink]

Show Tags

New post 11 Apr 2018, 09:46
Guess Strategy ( if time is less for computation )

Consider there were 5 yellow and 5 blue balls , the probality is 50% for each ball getting selected , hence the answer will somewhere hover around 1/2 , D is closer to 1/2 hence Guess D and move on.
GMAT Club Bot
Re: A jar of marbles contains 6 yellow and 4 blue &nbs [#permalink] 11 Apr 2018, 09:46
Display posts from previous: Sort by

A Jar of marble contains 6 yellow and 4 blue marbles, and no other mar

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


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.