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:

75% (01:03) correct 25% (01:31) wrong based on 100 sessions

HideShow timer Statistics

Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

Show Tags

13 May 2017, 19:21

1

This post received KUDOS

Bunuel wrote:

Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 2/11

D. 3/11

E. 1/3

To pick randomly 2 cans out of his refrigerator, there are \(12C2=66\) different ways. To pick randomly 2 cans out of his refrigerator and these 2 cans are cola and beer, there are \(4 \times 4 = 16\) different ways.

Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 2/11

D. 3/11

E. 1/3

Probability = \(\frac{Favourable outcomes}{Total outcomes}\) = 1- (\(\frac{Unfavourable outcomes}{Total outcomes}\))

Ways to pull one can of Cola = 4C1 = 4 Ways to pull one can of Root Beer = 4C1 = 4

Favourable outcome = 4C1 * 4C1 = 4*4 = 16

Total Outcomes = total ways to pull 2 cans out of total 12 cans = 12C2 = 66

Probability = 16/66 = 8/33

Bunuel Please check the options as there seems to be some mistake
_________________

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

Re: Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, an [#permalink]

Show Tags

14 May 2017, 13:13

1

This post received KUDOS

I solved it via two methods ==> Method (1)---> which I generally use --> Probability of the event => 4/12 *4/11 + 4/12 *4/11 => 8/33

Since it was not mentioned in the options ==> I moved to a slightly different method -->

Method 2--->

Total cases --> (C1,C2)(C1,C3)(C1,C4)(C1,B1)...(C1,B4)(C1,G1)...(C1,G4)=> 11 cases for each item Total cases => 11*12 Favourable cases => 4+4+4+4+4+4+4+4=> 32 cases

Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 8/33

D. 3/11

E. 1/3

Let’s first determine the number of ways Jerome can pull 1 can of root beer and 1 can of cola from the fridge. The number of ways to pull the cola is 4C1 = 4 and the number of ways to pull the root beer is 4C1 = 4.

The total number of ways to pull 2 cans from the fridge, with no restrictions, is 12C2 = (12 x 11)/2! = 66 ways.

Thus, the probability is (4 x 4)/66 = 16/66 = 8/33.

Answer: C
_________________

Scott Woodbury-Stewart Founder and CEO

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

Jerome's refrigerator contains 4 cans of cola, 4 cans of root beer, and 4 cans of ginger ale. If Jerome were to randomly pull two cans out of his refrigerator, what is the probability that he would pull out one can of cola and one can of root beer?

A. 1/11

B. 1/9

C. 8/33

D. 3/11

E. 1/3

P(one can of cola and one can of root beer) = P(1st can is cola AND 2nd can is root beer OR 1st can is root beer AND 2nd can is cola) = P(1st can is cola AND 2nd can is root beer) + P(1st can is root beer AND 2nd can is cola) = P(1st can is cola) x 2nd can is root beer) + P(1st can is root beer) x P(2nd can is cola) = 4/12 x 4/11 + 4/12 x 4/11 = 4/33 + 4/33 = 8/33 = C

We have 4 cans each of: cola, root beer and ginger ale - which gives us a total of 12 cans. We're asked for the probability of pulling 2 cans and getting 1 cola and one root beer. The math behind this question can be approached in a couple of different ways.

1st can = cola OR root beer = 8/12

Once we pull that first can, there will then be 11 cans remaining and we'll need to pull the OTHER type (re: if we pulled cola 1st, then we would need to pull root beer 2nd - and vice versa).

2nd CAT = the other type = 4/11

Thus, the probability = (8/12)(4/11) = 32/132 = 8/33