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

It is currently 18 Aug 2018, 11:36

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 miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal

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

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47983
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 08 Jul 2015, 03:10
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

60% (01:07) correct 40% (01:54) wrong based on 172 sessions

HideShow timer Statistics

A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4

Kudos for a correct solution.

_________________

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

Most Helpful Expert Reply
SVP
SVP
User avatar
P
Joined: 08 Jul 2010
Posts: 2139
Location: India
GMAT: INSIGHT
WE: Education (Education)
Reviews Badge
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 08 Jul 2015, 03:38
6
2
Bunuel wrote:
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4

Kudos for a correct solution.


Total balls in the Machine = 7+5+4 = 16

METHOD-1

Probability of First ball to be Blue = 7/(7+5+4) = 7/16
Probability of Second ball to be Green = 5/(15) = 5/15 [15 balls remaining after one being taken out in Step 1]
Probability of Third ball to be red = 4/(14) = 4/14 [14 balls remaining after Two being taken out in Step 1&2]

But this can happen in any order so the orders in which it can happen = 3!

i.e. Final Probability that machine dispenses one gumball of each color = (7/16)*(5/15)*(4/14)*3! = 1/4

Answer: Option E

METHOD-2

The total ways in which three balls can be Picked out of a total of 16 balls (with orders) = 16*15*14 = 840

The ways in which three balls of different colors can be taken out (with orders) = 7*5*4*3! = 3360

Probability = Favourable Outcome / Total Outcomes

i.e Probability = 840 / 3360 = 1/4

Answer: Option E

METHOD-3

The total ways in which three balls can be Picked out of a total of 16 balls (without orders) = 16C3 = 560

The ways in which three balls of different colors can be taken out (with orders) = 7C1 * 5C1 *4C1 = 140

Probability = Favourable Outcome / Total Outcomes

i.e Probability = 140 / 560 = 1/4

Answer: Option E
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

General Discussion
Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post Updated on: 13 Jul 2015, 12:31
Bunuel wrote:
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4

Kudos for a correct solution.

Probability of selecting a red ball = P(R) = 4/16

Probability of selecting a green ball = P(G) = 5/15

Probability of selecting a blue ball = P(B) = 7/14

Total number of ways we can select the RBG combination =3!

Thus the total probability is = (4/16)*(5/15)*(7/14)*3! = 1/4. Thus E is the correct answer.

Originally posted by ENGRTOMBA2018 on 08 Jul 2015, 04:17.
Last edited by ENGRTOMBA2018 on 13 Jul 2015, 12:31, edited 1 time in total.
Senior Manager
Senior Manager
User avatar
B
Joined: 28 Jun 2015
Posts: 295
Concentration: Finance
GPA: 3.5
Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 08 Jul 2015, 07:21
1
Total Events = 16C3 = 16*15*14/1*2*3 = 560

Probability = [7C1 * 5C1 * 4C1]/ 16C3 = 7 * 5 * 4/560 = 140/560 = 1/4. Ans (E).
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

Queens MBA Thread Master
avatar
Joined: 24 Oct 2012
Posts: 181
Concentration: Leadership, General Management
Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 08 Jul 2015, 21:49
1
Bunuel wrote:
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4

Kudos for a correct solution.


Probability equation can be written as
P(Differen color Balls = ( Number or ways Blue can be selected * Number or ways Green can be selected * Number or ways Red can be selected ) / total Number of ways 3 balls can be selected out of all.

Total Number of Ways in which 3 balls can be selected out of 16 = 16C3
Total Number of ways in which 1 blue ball can be selected out of 7 blue balls = 7C1 = 7
Total Number of ways in which 1 Green ball can be selected out of 5 Green balls = 5C1 = 5
Total Number of ways in which 1 Red ball can be selected out of 4 Red balls = 4C1 = 4

Hence Required probability = (16*7*5) / 16C3 = 1/4

Option E should be correct answer.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47983
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 13 Jul 2015, 12:14
Bunuel wrote:
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Consider one specific case: blue first, then green, then red. By the domino-effect rule, the probability of this case is:

\(\frac{7 \ blue}{16 \ total}*\frac{5 \ green}{15 \ total}*\frac{4 red}{14 \ total}=\frac{1}{24}\).

Now consider another case: green first, then red, then blue. The probability of this case is \(\frac{5 \ green}{16 \ total}*\frac{4 red}{15 \ total}*\frac{7 \ blue}{14 \ total}=\frac{1}{24}\). Notice that all we have done is swap around the numerators. We get the same final probability! This is no accident; the order in which the balls come out does not matter.

Because the three desired gum balls can come out in any order, there are 3! = 6 different cases. All of these cases must have the same probability. Therefore, the overall probability is 6*1/24 = 1/4.

In general, when you have a symmetrical problem with multiple equivalent cases, calculate the probability of one case (often by using the domino-effect rule). Then multiply by the number of cases. Use combinatorics to calculate the number of cases, if necessary.

Remember that when you apply a symmetry argument, the situation must truly be symmetrical. In the case above, if you swapped the order of "red" and "green" emerging from the gumball machine, nothing would change about the problem. As a result, we can use symmetry to simplify the computation.
_________________

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

SVP
SVP
User avatar
P
Joined: 08 Jul 2010
Posts: 2139
Location: India
GMAT: INSIGHT
WE: Education (Education)
Reviews Badge
Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 14 Jul 2015, 05:00
1
Bunuel wrote:
Bunuel wrote:
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Consider one specific case: blue first, then green, then red. By the domino-effect rule, the probability of this case is:

\(\frac{7 \ blue}{16 \ total}*\frac{5 \ green}{15 \ total}*\frac{4 red}{15\ total}=\frac{1}{24}\).

Now consider another case: green first, then red, then blue. The probability of this case is \(\frac{5 \ green}{16 \ total}*\frac{4 red}{15 \ total}*\frac{7 \ blue}{14 \ total}=\frac{1}{24}\). Notice that all we have done is swap around the numerators. We get the same final probability! This is no accident; the order in which the balls come out does not matter.

Because the three desired gum balls can come out in any order, there are 3! = 6 different cases. All of these cases must have the same probability. Therefore, the overall probability is 6*1/24 = 1/3.

In general, when you have a symmetrical problem with multiple equivalent cases, calculate the probability of one case (often by using the domino-effect rule). Then multiply by the number of cases. Use combinatorics to calculate the number of cases, if necessary.

Remember that when you apply a symmetry argument, the situation must truly be symmetrical. In the case above, if you swapped the order of "red" and "green" emerging from the gumball machine, nothing would change about the problem. As a result, we can use symmetry to simplify the computation.


Bunuel:

The Highlighted parts are representing mistakes

The first Highlighted part 15 should have been 14
and
The Second Highlighted part 6*1/24 = 1/3 should have been 6*1/24 = 1/4
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47983
Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 14 Jul 2015, 05:05
GMATinsight wrote:
Bunuel wrote:
Bunuel wrote:
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Consider one specific case: blue first, then green, then red. By the domino-effect rule, the probability of this case is:

\(\frac{7 \ blue}{16 \ total}*\frac{5 \ green}{15 \ total}*\frac{4 red}{15\ total}=\frac{1}{24}\).

Now consider another case: green first, then red, then blue. The probability of this case is \(\frac{5 \ green}{16 \ total}*\frac{4 red}{15 \ total}*\frac{7 \ blue}{14 \ total}=\frac{1}{24}\). Notice that all we have done is swap around the numerators. We get the same final probability! This is no accident; the order in which the balls come out does not matter.

Because the three desired gum balls can come out in any order, there are 3! = 6 different cases. All of these cases must have the same probability. Therefore, the overall probability is 6*1/24 = 1/3.

In general, when you have a symmetrical problem with multiple equivalent cases, calculate the probability of one case (often by using the domino-effect rule). Then multiply by the number of cases. Use combinatorics to calculate the number of cases, if necessary.

Remember that when you apply a symmetry argument, the situation must truly be symmetrical. In the case above, if you swapped the order of "red" and "green" emerging from the gumball machine, nothing would change about the problem. As a result, we can use symmetry to simplify the computation.


Bunuel:

The Highlighted parts are representing mistakes

The first Highlighted part 15 should have been 14
and
The Second Highlighted part 6*1/24 = 1/3 should have been 6*1/24 = 1/4


Edited the typos. Thank you.
_________________

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

Intern
Intern
avatar
B
Joined: 20 Oct 2015
Posts: 3
Location: Indonesia
Concentration: General Management, Strategy
GMAT 1: 650 Q44 V36
GMAT 2: 710 Q48 V39
WE: Consulting (Consulting)
Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 12 Mar 2017, 09:28
GMATinsight wrote:
Bunuel wrote:
A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?

A. 35/1024
B. 1/24
C. 1/8
D. 105/512
E. 1/4

Kudos for a correct solution.


Total balls in the Machine = 7+5+4 = 16

METHOD-1

Probability of First ball to be Blue = 7/(7+5+4) = 7/16
Probability of Second ball to be Green = 5/(15) = 5/15 [15 balls remaining after one being taken out in Step 1]
Probability of Third ball to be red = 4/(14) = 4/14 [14 balls remaining after Two being taken out in Step 1&2]

But this can happen in any order so the orders in which it can happen = 3!

i.e. Final Probability that machine dispenses one gumball of each color = (7/16)*(5/15)*(4/14)*3! = 1/4

Answer: Option E

METHOD-2

The total ways in which three balls can be Picked out of a total of 16 balls (with orders) = 16*15*14 = 840

The ways in which three balls of different colors can be taken out (with orders) = 7*5*4*3! = 3360

Probability = Favourable Outcome / Total Outcomes

i.e Probability = 840 / 3360 = 1/4

Answer: Option E

METHOD-3

The total ways in which three balls can be Picked out of a total of 16 balls (without orders) = 16C3 = 560

The ways in which three balls of different colors can be taken out (with orders) = 7C1 * 5C1 *4C1 = 140

Probability = Favourable Outcome / Total Outcomes

i.e Probability = 140 / 560 = 1/4

Answer: Option E


Can anyone please help me to understand why the orders of the balls considered important when calculating the number of desired outcomes?
Thank you in advance.

Regards,
Nathan
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 7759
Premium Member
Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal  [#permalink]

Show Tags

New post 06 Apr 2018, 10:03
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 Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal &nbs [#permalink] 06 Apr 2018, 10:03
Display posts from previous: Sort by

A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal

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

Events & Promotions

PREV
NEXT


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®.