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

It is currently 24 Apr 2015, 22:37

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A box contains 10 light bulbs, fewer than half of which are

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
User avatar
Joined: 06 Apr 2010
Posts: 144
Followers: 3

Kudos [?]: 198 [0], given: 15

Reviews Badge
A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 28 Aug 2010, 01:54
7
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

69% (02:29) correct 31% (02:17) wrong based on 416 sessions
A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.
[Reveal] Spoiler: OA
Expert Post
2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 27053
Followers: 4183

Kudos [?]: 40303 [2] , given: 5420

Re: Defective Bulb Probability [#permalink] New post 28 Aug 2010, 05:09
2
This post received
KUDOS
Expert's post
6
This post was
BOOKMARKED
udaymathapati wrote:
A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.


Given: \(bulbs=10\) and \(defective=n<5\). Question: \(n=?\)

(1) The probability that the two bulbs to be drawn will be defective is 1/15 --> clearly sufficient, as probability, \(p\), of drawing 2 defective bulbs out of total 10 bulbs, obviously depends on # of defective bulbs, \(n\), so we can calculate uniques value of \(n\) if we are given \(p\).

To show how it can be done: \(\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}\) --> \(n(n-1)=6\) --> \(n=3\) or \(n=-2\) (not a valid solution as \(n\) represents # of defective bulbs and can not be negative). Sufficient.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15 --> also sufficient, but a little bit trickier: if it were 3 defective and 7 good bulbs OR 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases (symmetric distribution), so info about the probability, 7/15, of drawing one defective and one good bulb would give us 2 values of \(n\) one less than 5 and another more than 5 (their sum would be 10), but as we are given that \(n<5\), we can stiil get unique value of \(n\) which is less than 5.

To show how it can be done: \(2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\) --> \(n(10-n)=21\) --> \(n=3\) or \(n=7\) (not a valid solution as \(n<5\)). Sufficient.

Answer: D.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

SVP
SVP
avatar
Joined: 17 Feb 2010
Posts: 1560
Followers: 13

Kudos [?]: 275 [0], given: 6

Re: Defective Bulb Probability [#permalink] New post 29 Aug 2010, 18:52
Bunuel,

I did not understand the below part in option 2. How you got '2' in the equation. I know n < 5 but why do we need to multiply by '2'

\(2 * n/10 * (n-1)/9 = 1/15\)

Thanks
Intern
Intern
avatar
Joined: 05 Nov 2009
Posts: 32
Followers: 0

Kudos [?]: 4 [0], given: 3

GMAT ToolKit User
Re: Defective Bulb Probability [#permalink] New post 29 Aug 2010, 19:00
Seekmba,

There are 2 ways to draw 2 bulbs in which 1 works and 1 is defective.
SVP
SVP
avatar
Joined: 17 Feb 2010
Posts: 1560
Followers: 13

Kudos [?]: 275 [0], given: 6

Re: Defective Bulb Probability [#permalink] New post 29 Aug 2010, 20:07
But that is why we have

\(n/10 * (n-1)/9 = 1/15\).

I am sorry but I still dont understand why multiply by 2.
1 KUDOS received
Director
Director
avatar
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
Posts: 691
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Followers: 14

Kudos [?]: 86 [1] , given: 15

Re: Defective Bulb Probability [#permalink] New post 29 Aug 2010, 20:54
1
This post received
KUDOS
Think of it like this:
Can select 2 bulbs out of 10 in 10C2 ways

1 defective and 1 ok bulb in N C1 x 10-N C1

So probability is 2 x N x 10-N / (9 x 10) = 7/15
_________________

Consider kudos, they are good for health

Expert Post
3 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 27053
Followers: 4183

Kudos [?]: 40303 [3] , given: 5420

A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 30 Aug 2010, 07:03
3
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
seekmba wrote:
Bunuel,

I did not understand the below part in option 2. How you got '2' in the equation. I know n < 5 but why do we need to multiply by '2'

\(2 * n/10 * (n-1)/9 = 1/15\)

Thanks


The probability that one of the bulbs to be drawn will be defective and the other will not be defective is the sum of the probabilities of 2 events: the first one is defective and the second is not PLUS the first is not defective and the second one is defective = \(\frac{n}{10}*\frac{10-n}{9}+\frac{10-n}{10}*\frac{n}{9}=2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\).
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

Manager
Manager
avatar
Joined: 16 Mar 2010
Posts: 190
Followers: 2

Kudos [?]: 62 [0], given: 9

Re: Defective Bulb Probability [#permalink] New post 31 Aug 2010, 03:32
Bunuel Rocks.....
Intern
Intern
avatar
Joined: 03 Feb 2011
Posts: 2
Followers: 0

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

Re: Defective Bulb Probability [#permalink] New post 03 Feb 2011, 21:46
Yes, that it was only for two ways. However, we must see the option into the works of Bunuel. That it was the possible way to check and see the probability of the light bulb.
2 KUDOS received
Math Forum Moderator
avatar
Joined: 20 Dec 2010
Posts: 2035
Followers: 135

Kudos [?]: 1098 [2] , given: 376

Re: Defective Bulb Probability [#permalink] New post 03 Feb 2011, 23:54
2
This post received
KUDOS
Defective bulbs: n
Non-defective bulbs: 10-n

and 0<=n<5.

Two bulbs can be drawn from a lot of 10 in
\(C^{10}_{2} = \frac{10!}{2!8!}=\frac{(10*9)}{2}=45\)
ways.


(1) Both defective:

Probability(Pick both defective bulbs) = Number of favorable choices/Number of total choices

Number of favorable choices : If we draw both defective bulbs
i.e. {draw 2 balls out of n defective balls} * {0 balls out of (10-n) balls}
The number of ways that can be done is:

\(C^{n}_{2} * C^{(10-n)}_{0}\)

\(=> \frac{n!}{(n-2)!2!} * 1\)

\(\frac{n(n-1)(n-2)!}{(n-2)!2!}=\frac{n(n-1)}{2}\)

Number of favorable choices: \(\frac{n(n-1)}{2}\)

Number of total choices: 45

Probability(Pick both defective bulbs) = \(\frac{n(n-1)}{(2*45)}\)

Given, Probability(Pick both defective bulbs) = 1/15

\(\frac{n(n-1)}{90} = \frac{1}{15}\)
\(n(n-1) = 6\)
\(n^2-n-6=0\)

Solving the quadratic equation;

n=3, n=-2

Since,
0<=n<5
n = 3

Sufficient.


(2) One defective and one non-defective:

Probability(One defective and one non-defective) = Number of favorable choices/Number of total choices

Number of favorable choices : If we draw exactly one defective and one non-defective bulb
i.e. {draw 1 balls out of n defective balls} * {1 balls out of (10-n) balls}
The number of ways that can be done is:

\(C^{n}_{1} * C^{(10-n)}_{1}\)

\(=> \frac{n!}{(n-1)!1!} * \frac{(10-n)!}{(10-n-1)!1!}\)

\(=> \frac{n(n-1)!}{(n-1)!1!} * \frac{(10-n)(10-n-1)!}{(10-n-1)!1!}\)

\(n(10-n)\)

Number of favorable choices: \(n(10-n)\)

Number of total choices: 45

Probability(One defective and one non-defective) = \(\frac{n(10-n)}{45}\)

Given, Probability(One defective and one non-defective) = 7/15


\(\frac{n(10-n)}{45} = \frac{7}{15}\)
\(n(10-n) = 21\)
\(n^2-10n+21=0\)

Solving the quadratic equation gives:
n=3 and n=7

Since,
0<=n<5
n = 3

Sufficient.

Ans: "D"
_________________

~fluke

GMAT Club Premium Membership - big benefits and savings

Manager
Manager
avatar
Joined: 14 Feb 2012
Posts: 228
Followers: 1

Kudos [?]: 98 [0], given: 7

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 08 May 2012, 04:37
Hi Bunuel ,
I am not very sure about the concept of symmetric probability , and when to use it ?
Can you please explain with an example.
Thanks
_________________

The Best Way to Keep me ON is to give Me KUDOS !!!
If you Like My posts please Consider giving Kudos

Shikhar

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 27053
Followers: 4183

Kudos [?]: 40303 [0], given: 5420

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 09 May 2012, 00:33
Expert's post
shikhar wrote:
Hi Bunuel ,
I am not very sure about the concept of symmetric probability , and when to use it ?
Can you please explain with an example.
Thanks


Good question about this concept: mary-and-joe-126407.html

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 27053
Followers: 4183

Kudos [?]: 40303 [0], given: 5420

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 07 Jun 2013, 05:11
Expert's post
1
This post was
BOOKMARKED
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

3 KUDOS received
Manager
Manager
avatar
Joined: 28 Feb 2012
Posts: 115
Concentration: Strategy, International Business
Schools: INSEAD Jan '13
GPA: 3.9
WE: Marketing (Other)
Followers: 0

Kudos [?]: 23 [3] , given: 17

GMAT ToolKit User
Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 08 Jun 2013, 01:08
3
This post received
KUDOS
udaymathapati wrote:
A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.



As an alternative way i suggest to use plug in method.

(1) since fewer than 5 bulbs are defective we are limited to choices 4,3,2 or 1. So max 4 numbers to plug, not many. First lets take 4 defective bulbs. (4/10)*(3/9)=12/90=2/15 a little bigger than 1/15, so the number of defective bulbs must be 3, but lets check it. (3/10)*(2/9)=6/90=1/15. Yes we could find the number of defective bulbs using the statement 1 - Sufficient.

(2) we can use the same logic here as well. The only difference is that here we need to look for one defective and one not defective sequence and multiple it to 2, because the sequence could start with defective as well as with non-defective bulb. Lets take 4 as the number of defective bulbs: (4/10)*(7/9)=24/90=8/30 after multiplying to 2 we have 8/15 not the ration we are looking for. Lets take 3 as the number of defective bulbs: (3/10)*(7/9)=21/90=7/30 after multiplying it to 2 we have 7/15. This is the ratio that we were looking for, so the statement 2 is sufficient.

Since both statements are sufficient on their own the answer is D.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 2025
Concentration: Finance
GMAT 1: 710 Q48 V39
Followers: 24

Kudos [?]: 290 [0], given: 354

GMAT ToolKit User
Re: Defective Bulb Probability [#permalink] New post 01 Apr 2014, 05:19
Bunuel wrote:
udaymathapati wrote:
A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.


Given: \(bulbs=10\) and \(defective=n<5\). Question: \(n=?\)

(1) The probability that the two bulbs to be drawn will be defective is 1/15 --> clearly sufficient, as probability, \(p\), of drawing 2 defective bulbs out of total 10 bulbs, obviously depends on # of defective bulbs, \(n\), so we can calculate uniques value of \(n\) if we are given \(p\).

To show how it can be done: \(\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}\) --> \(n(n-1)=6\) --> \(n=3\) or \(n=-2\) (not a valid solution as \(n\) represents # of defective bulbs and can not be negative). Sufficient.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15 --> also sufficient, but a little bit trickier: if it were 3 defective and 7 good bulbs OR 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases (symmetric distribution), so info about the probability, 7/15, of drawing one defective and one good bulb would give us 2 values of \(n\) one less than 5 and another more than 5 (their sum would be 10), but as we are given that \(n<5\), we can stiil get unique value of \(n\) which is less than 5.

To show how it can be done: \(2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\) --> \(n(10-n)=21\) --> \(n=3\) or \(n=7\) (not a valid solution as \(n<5\)). Sufficient.

Answer: D.

Hope it's clear.



In the second statement, you didn't subtract the defective bulb that was taken first. Was this because the question says that they are taken simultaneously?

Thanks for clarifying
Cheers
J
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 27053
Followers: 4183

Kudos [?]: 40303 [0], given: 5420

Re: Defective Bulb Probability [#permalink] New post 01 Apr 2014, 07:27
Expert's post
jlgdr wrote:
Bunuel wrote:
udaymathapati wrote:
A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.


Given: \(bulbs=10\) and \(defective=n<5\). Question: \(n=?\)

(1) The probability that the two bulbs to be drawn will be defective is 1/15 --> clearly sufficient, as probability, \(p\), of drawing 2 defective bulbs out of total 10 bulbs, obviously depends on # of defective bulbs, \(n\), so we can calculate uniques value of \(n\) if we are given \(p\).

To show how it can be done: \(\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}\) --> \(n(n-1)=6\) --> \(n=3\) or \(n=-2\) (not a valid solution as \(n\) represents # of defective bulbs and can not be negative). Sufficient.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15 --> also sufficient, but a little bit trickier: if it were 3 defective and 7 good bulbs OR 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases (symmetric distribution), so info about the probability, 7/15, of drawing one defective and one good bulb would give us 2 values of \(n\) one less than 5 and another more than 5 (their sum would be 10), but as we are given that \(n<5\), we can stiil get unique value of \(n\) which is less than 5.

To show how it can be done: \(2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\) --> \(n(10-n)=21\) --> \(n=3\) or \(n=7\) (not a valid solution as \(n<5\)). Sufficient.

Answer: D.

Hope it's clear.



In the second statement, you didn't subtract the defective bulb that was taken first. Was this because the question says that they are taken simultaneously?

Thanks for clarifying
Cheers
J


Not sure I understand what you mean...

Anyway, mathematically the probability of picking two balls simultaneously, or picking them one at a time (without replacement) is the same.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

Manager
Manager
User avatar
Joined: 01 Oct 2013
Posts: 95
Schools: Haas '16, AGSM '16
Followers: 0

Kudos [?]: 1 [0], given: 13

GMAT ToolKit User
Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 22 Jul 2014, 08:57
D is correct

b=10 , d=n<5 , Select 2

The total possibilities to select 2 out of 10 bulbs: 10C2 = 45
.
Total possibilities to select 2 out of n defective bulbs: nC2 = 1/2 * n(n-1)
.
(1) => (1/2 * n(n-1))/45 =1/5 => n={3;-2} => (1) sufficient
.
Total possibilities to either select 2 out of 10-n bulbs or n defective bulbs: nC2 + (10-n)C2 = 1/2 * n(n-1) + 1/2 * (10-n)(9-n)
.
(2) => the probability to either select 2 out of 10-n bulbs or n defective bulbs is 8/15
we have the equation:
(1/2 * n(n-1) + 1/2 * (10-n)(9-n))/45 = 8/15
=> n = {3;7} => (2) sufficient
Intern
Intern
avatar
Joined: 25 Sep 2014
Posts: 7
Followers: 0

Kudos [?]: 0 [0], given: 1

A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 11 Dec 2014, 02:49
Hi Bunuel, I think I have the same question as jlgdr. (In the second statement, you didn't subtract the defective bulb that was taken first. Was this because the question says that they are taken simultaneously?)

In other words, why is the equation 2x(n/10)x((10-n)/9) and not 2x(n/10)x((10-n-1)/9). Is this because they are taken simultaneously?
Or is this because the second term is (9-(n-1))/9 = (9-n+1)/9 =(10-n)/9 as there is one "n" less to deduct after pulling it out in the first draw?

Appreciate your help,
Many thanks

Last edited by Parco on 11 Dec 2014, 04:17, edited 1 time in total.
Intern
Intern
User avatar
Joined: 21 Feb 2012
Posts: 27
Followers: 0

Kudos [?]: 9 [0], given: 54

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 11 Dec 2014, 03:49
as per question n ranges from 0 to 4 and is an integer
statement 1
nC2/10C2 = 1/15
this gives us a quadratic equation with solutions as 3 & -2. 3 is valid. Hence sufficient
statement 2
nC1*(10-n)C1/10C2 = 7/15
this also gives us a quadratic equation with 2 +ve solns of 3 & 7. But as per the range of n it cannot be 7. So n=3. Hence sufficient.

Answer is D
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 27053
Followers: 4183

Kudos [?]: 40303 [0], given: 5420

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 11 Dec 2014, 05:51
Expert's post
Parco wrote:
Hi Bunuel, I think I have the same question as jlgdr. (In the second statement, you didn't subtract the defective bulb that was taken first. Was this because the question says that they are taken simultaneously?)

In other words, why is the equation 2x(n/10)x((10-n)/9) and not 2x(n/10)x((10-n-1)/9). Is this because they are taken simultaneously?
Or is this because the second term is (9-(n-1))/9 = (9-n+1)/9 =(10-n)/9 as there is one "n" less to deduct after pulling it out in the first draw?

Appreciate your help,
Many thanks


After 1 defective bulb is drawn there are total 9 bulbs left out of which n-1 are defective and 9 - (n-1) = 10 - n non-defective. So, the probability of drawing non-defective is (non-defective)/(total) = (10 - n)/9.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

Re: A box contains 10 light bulbs, fewer than half of which are   [#permalink] 11 Dec 2014, 05:51
    Similar topics Author Replies Last post
Similar
Topics:
A box contains 10 light bulbs, fewer than half of which are tenaman10 2 06 Jun 2009, 01:35
A box contains 10 light bulbs, fewer than half of which are sidbidus 3 30 Oct 2007, 09:46
A box contains 10 light bulbs, fewer than half of which are gluon 3 08 Sep 2007, 15:18
A box contains 10 light bulbs, fewer than half of which are sidbidus 2 30 Aug 2007, 07:58
A box contains 10 light bulbs, fewer than half of which are WinIT 6 12 Nov 2006, 08:40
Display posts from previous: Sort by

A box contains 10 light bulbs, fewer than half of which are

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

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®.