Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 30 Sep 2016, 08:24

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 12 Jun 2012
Posts: 42
Followers: 1

Kudos [?]: 28 [0], given: 28

A gumball machine contains 7 blue, 5 green, and 4 red [#permalink]

### Show Tags

22 Oct 2012, 01:22
2
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 4 sessions

### HideShow timer Statistics

A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Use the comibatorics method only!
_________________

Intern
Joined: 12 Jun 2012
Posts: 42
Followers: 1

Kudos [?]: 28 [0], given: 28

Re: A gumball machine contains... [#permalink]

### Show Tags

22 Oct 2012, 01:27
I tried to solve using combinatorics but I was wrong - answer below.

I thought there is 16 balls in total and if we select three from 16, we have 16!/3!13! ways to choose three balls.

BGR is one of each and there is 6 ways of organising

we have 1C7x 1C5 x 1C4 x 6 / number of ways to choose three balls

so we have 7 x 5 x 4 x 6 / 16 x 15 x 14 x 3 x 2

Simplifying the sixes 7 x 5 x 4 / 16 x 15 x 14

simplifying 7 and 14 1 x 5 x 4 / 16 x 15 x 2

Simplifying 5 and 15 1 x 1 x 4 / 16 x 3 x 2

Simplifying 4, 16 1 x 1 x 1 / 4 x 3 x 2

However

Using simple probabilities

6x 7/16 x 5/15 x 4/14 = 1/4
_________________

Moderator
Joined: 02 Jul 2012
Posts: 1231
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Followers: 110

Kudos [?]: 1289 [1] , given: 116

Re: A gumball machine contains... [#permalink]

### Show Tags

22 Oct 2012, 02:35
1
KUDOS
Total No. of ways to select 3 out of 16 is = 16C3

No. Of ways to select one blue ball = 7C1
No. Of ways to select one green ball = 5C1
No. Of ways to select one red ball = 4C1

So. total no. of ways for favourable outcome = 7*5*4 = 140

So Probability = 140/560 = 1/4

When using simple probability, after choosing the first ball, there are only 15 balls to choose from and one less of the color that has already been chosen and after selecting the second, there are only 14 balls to choose from and one less of the color that has already been chosen and so on. Hence order has to be taken into account.

However while using combinations, the order does not matter as we are selecting one ball from one color only each time.
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 92

Kudos [?]: 834 [0], given: 43

Re: A gumball machine contains... [#permalink]

### Show Tags

22 Oct 2012, 03:49
jordanshl wrote:
I tried to solve using combinatorics but I was wrong - answer below.

I thought there is 16 balls in total and if we select three from 16, we have 16!/3!13! ways to choose three balls.

BGR is one of each and there is 6 ways of organising

we have 1C7x 1C5 x 1C4 x 6 / number of ways to choose three balls

so we have 7 x 5 x 4 x 6 / 16 x 15 x 14 x 3 x 2

Simplifying the sixes 7 x 5 x 4 / 16 x 15 x 14

simplifying 7 and 14 1 x 5 x 4 / 16 x 15 x 2

Simplifying 5 and 15 1 x 1 x 4 / 16 x 3 x 2

Simplifying 4, 16 1 x 1 x 1 / 4 x 3 x 2

However

Using simple probabilities

6x 7/16 x 5/15 x 4/14 = 1/4

You did a computational mistake, the 3! from the denominator of 16C3 should go up to the numerator.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Math Expert
Joined: 02 Sep 2009
Posts: 34902
Followers: 6505

Kudos [?]: 83126 [1] , given: 10143

Re: A gumball machine contains... [#permalink]

### Show Tags

22 Oct 2012, 08:35
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
jordanshl wrote:
A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Use the comibatorics method only!

We need to find the probability of BGR (a marble of each color).

Probability approach:

$$P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}$$, we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).

Combination approach:

$$P(BGR)=\frac{C^1_7*C^1_5*C^1_4}{C^3_{16}}=\frac{1}{4}$$.

Hope it's clear.

_________________
Intern
Joined: 08 Jul 2011
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 5

Re: A gumball machine contains... [#permalink]

### Show Tags

09 Jul 2013, 08:05
Bunuel wrote:
jordanshl wrote:
A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Use the comibatorics method only!

We need to find the probability of BGR (a marble of each color).

Probability approach:

$$P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}$$, we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).

Combination approach:

$$P(BGR)=\frac{C^1_7*C^1_5*C^1_4}{C^3_{16}}=\frac{1}{4}$$.

Hope it's clear.

Is there a way to solve this using $$\frac{16!}{7!*5!*4!}$$ as the denominator?
Math Expert
Joined: 02 Sep 2009
Posts: 34902
Followers: 6505

Kudos [?]: 83126 [1] , given: 10143

Re: A gumball machine contains... [#permalink]

### Show Tags

10 Jul 2013, 00:14
1
KUDOS
Expert's post
elementbrdr wrote:
Bunuel wrote:
jordanshl wrote:
A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Use the comibatorics method only!

We need to find the probability of BGR (a marble of each color).

Probability approach:

$$P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}$$, we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).

Combination approach:

$$P(BGR)=\frac{C^1_7*C^1_5*C^1_4}{C^3_{16}}=\frac{1}{4}$$.

Hope it's clear.

Is there a way to solve this using $$\frac{16!}{7!*5!*4!}$$ as the denominator?

Yes.

Total # of outcomes: $$\frac{16!}{7!*5!*4!}$$ (the number of arrangements of the mables).

Favorable outcomes: we need the first three marbles to be BGR in any combination, so 3!. The remaining 13 marbles (6 blue, 4 green, and 3 red) can be arranged in 13!/(6!4!3!).

P = (Favorable)/(Total) = $$\frac{(3!\frac{13!}{6!4!3!})}{(\frac{16!}{7!*5!*4!})}=\frac{1}{4}$$.

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11757
Followers: 529

Kudos [?]: 149 [0], given: 0

Re: A gumball machine contains 7 blue, 5 green, and 4 red [#permalink]

### Show Tags

01 Sep 2016, 03:10
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.
_________________
Re: A gumball machine contains 7 blue, 5 green, and 4 red   [#permalink] 01 Sep 2016, 03:10
Similar topics Replies Last post
Similar
Topics:
1 In a particular gumball machine, there are 4 identical blue gumballs, 1 08 Apr 2016, 02:44
7 A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal 7 08 Jul 2015, 03:10
A deck of cards contain 2 red , 3 blue , 4 green cards. Three cards ar 8 31 Aug 2010, 13:23
1 A bag contains 6 red marbles,9 blue marbles,and 5 green marbles.You wi 2 25 Aug 2010, 19:22
2 A deck of 9 cards contains 2 red cards, 3 blue cards, and 4 green card 2 17 Jun 2010, 06:20
Display posts from previous: Sort by