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

It is currently 22 Oct 2018, 06:02

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

Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,

  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: 50039
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

Show Tags

New post 29 Sep 2016, 04:35
1
1
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

80% (01:57) correct 20% (01:45) wrong based on 119 sessions

HideShow timer Statistics

Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?

A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8

_________________

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

Retired Moderator
avatar
G
Joined: 26 Nov 2012
Posts: 593
Premium Member
Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

Show Tags

New post 29 Sep 2016, 04:48
1
Bunuel wrote:
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?

A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8


P(atleast ) = 1 - P(none)

Now we need to find none of the tv sets are broken.

We have 8 tvs all together..

First attempt only the fixed one = 6/8.

Second attempt only the fixed ones = 5/7...(since already one tv set is taken out)

both case = 6/8*5/7 = 15/28

P(atleast ) = 1 - P(none) => 1-15/28 = 13/28.

IMO option B
Senior Manager
Senior Manager
avatar
Joined: 06 Jun 2016
Posts: 259
Location: India
Concentration: Operations, Strategy
Schools: ISB '18 (D)
GMAT 1: 600 Q49 V23
GMAT 2: 680 Q49 V34
GPA: 3.9
Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

Show Tags

New post 29 Sep 2016, 05:34
total number of possible outcomes= 8C2= 28
Required is Total- (no of possibilities of selecting 2 repaired TVs)
28-6C2= 13
So probability= 13/28
Senior Manager
Senior Manager
User avatar
B
Status: Come! Fall in Love with Learning!
Joined: 05 Jan 2017
Posts: 472
Location: India
Premium Member
Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

Show Tags

New post 27 Feb 2017, 05:15
P(atleast one is broken) = 1- P(both fixed)
= 1 - 6C2/8C2
= 1- 15/28
= 13/28. Option B
_________________

GMAT Mentors
Image

Target Test Prep Representative
User avatar
P
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3907
Location: United States (CA)
Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

Show Tags

New post 01 Mar 2017, 18:11
1
Bunuel wrote:
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?

A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8


We are given that there are 2 broken TV sets and 6 fixed TV sets and need to determine the probability that when 2 TV sets are selected, at least 1 is broken. Recall that the phrase “at least one” means “one or more.”

We can use the following formula:

1 = P(at least 1 set is broken) + P(no sets are broken)

Thus:

P(at least 1 set is broken) = 1 - P(no sets are broken)

P(no sets are broken) = 6/8 x 5/7 = 3/4 x 5/7 = 15/28

P(at least 1 set is broken) = 1 - 15/28 = 13/28

Answer: B
_________________

Scott Woodbury-Stewart
Founder and CEO

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

CEO
CEO
User avatar
D
Joined: 12 Sep 2015
Posts: 3024
Location: Canada
Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,  [#permalink]

Show Tags

New post 08 Sep 2018, 08:46
Top Contributor
Bunuel wrote:
Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?

A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8


There are 8 TV's in total
2 are broken
6 are fixed

We want to find P(at least one TV is broken)
When it comes to probability questions involving at least, it's often best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 broken TV) = 1 - P(not getting at least 1 broken TV)
What does it mean to not get at least 1 broken TV? It means getting zero broken TVs.

So, we can write: P(getting at least 1 broken TV) = 1 - P(getting zero broken TVs)
1 - P(getting two FIXED TVs)

P(getting two FIXED TVs)
P(getting two FIXED TVs) = P(1st TV fixed and 2nd TV is fixed)
= P(1st TV fixed) x P(2nd TV is fixed)
= 6/8 x 5/7
= 30/56
= 15/28

So, P(getting at least 1 broken TV) = 1 - P(not getting at least 1 broken TV)
= 1 - 15/28
= 13/28

Answer: B

Cheers,
Brent
_________________

Brent Hanneson – GMATPrepNow.com
Image
Sign up for our free Question of the Day emails

GMAT Club Bot
Re: Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, &nbs [#permalink] 08 Sep 2018, 08:46
Display posts from previous: Sort by

Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store,

  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®.