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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

22 Oct 2012, 01:22

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct
0% (00:00) wrong based on 6 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!
_________________

If you find my post helpful, please GIVE ME SOME KUDOS!

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.

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

Answer would be 1/24

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. So your answer would be correct (1/24) x 6 =1/4.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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!).

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!).

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!).

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) = \(\frac{(3!\frac{13!}{6!4!3!})}{(\frac{16!}{7!*5!*4!})}=\frac{1}{4}\).

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.
_________________

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...