Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 06 Apr 2010
Posts: 135

A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
28 Aug 2010, 02:54
3
This post received KUDOS
17
This post was BOOKMARKED
Question Stats:
73% (01:35) correct 27% (02:18) wrong based on 822 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 45305

Re: Defective Bulb Probability [#permalink]
Show Tags
28 Aug 2010, 06:09
7
This post received KUDOS
Expert's post
18
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: \(\frac{n}{10}*\frac{n1}{9}=\frac{1}{15}\) > \(n(n1)=6\) > \(n=3\) 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: \(2*\frac{n}{10}*\frac{10n}{9}=\frac{7}{15}\) > \(n(10n)=21\) > \(n=3\) or \(n=7\) (not a valid solution as \(n<5\)). Sufficient. Answer: D. Hope it's clear.
_________________
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? Extrahard Quant Tests with Brilliant Analytics



VP
Joined: 17 Feb 2010
Posts: 1310

Re: Defective Bulb Probability [#permalink]
Show Tags
29 Aug 2010, 19:52
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'
\(2 * n/10 * (n1)/9 = 1/15\)
Thanks



Intern
Joined: 05 Nov 2009
Posts: 31

Re: Defective Bulb Probability [#permalink]
Show Tags
29 Aug 2010, 20:00
1
This post received KUDOS
Seekmba,
There are 2 ways to draw 2 bulbs in which 1 works and 1 is defective.



VP
Joined: 17 Feb 2010
Posts: 1310

Re: Defective Bulb Probability [#permalink]
Show Tags
29 Aug 2010, 21:07
But that is why we have
\(n/10 * (n1)/9 = 1/15\).
I am sorry but I still dont understand why multiply by 2.



Director
Status: Apply  Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 646
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas

Re: Defective Bulb Probability [#permalink]
Show Tags
29 Aug 2010, 21:54
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 10N C1 So probability is 2 x N x 10N / (9 x 10) = 7/15
_________________
Consider kudos, they are good for health



Math Expert
Joined: 02 Sep 2009
Posts: 45305

A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
30 Aug 2010, 08:03
6
This post received KUDOS
Expert's post
3
This post was BOOKMARKED



Manager
Joined: 17 Mar 2010
Posts: 156

Re: Defective Bulb Probability [#permalink]
Show Tags
31 Aug 2010, 04:32
Bunuel Rocks.....



Intern
Joined: 03 Feb 2011
Posts: 2

Re: Defective Bulb Probability [#permalink]
Show Tags
03 Feb 2011, 22:46
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.



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1903

Re: Defective Bulb Probability [#permalink]
Show Tags
04 Feb 2011, 00:54
3
This post received KUDOS
Defective bulbs: n Nondefective bulbs: 10n 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 (10n) balls} The number of ways that can be done is: \(C^{n}_{2} * C^{(10n)}_{0}\) \(=> \frac{n!}{(n2)!2!} * 1\) \(\frac{n(n1)(n2)!}{(n2)!2!}=\frac{n(n1)}{2}\) Number of favorable choices: \(\frac{n(n1)}{2}\) Number of total choices: 45 Probability(Pick both defective bulbs) = \(\frac{n(n1)}{(2*45)}\) Given, Probability(Pick both defective bulbs) = 1/15 \(\frac{n(n1)}{90} = \frac{1}{15}\) \(n(n1) = 6\) \(n^2n6=0\) Solving the quadratic equation; n=3, n=2 Since, 0<=n<5 n = 3 Sufficient. (2) One defective and one nondefective: Probability(One defective and one nondefective) = Number of favorable choices/Number of total choices Number of favorable choices : If we draw exactly one defective and one nondefective bulb i.e. {draw 1 balls out of n defective balls} * {1 balls out of (10n) balls} The number of ways that can be done is: \(C^{n}_{1} * C^{(10n)}_{1}\) \(=> \frac{n!}{(n1)!1!} * \frac{(10n)!}{(10n1)!1!}\) \(=> \frac{n(n1)!}{(n1)!1!} * \frac{(10n)(10n1)!}{(10n1)!1!}\) \(n(10n)\) Number of favorable choices: \(n(10n)\) Number of total choices: 45 Probability(One defective and one nondefective) = \(\frac{n(10n)}{45}\) Given, Probability(One defective and one nondefective) = 7/15 \(\frac{n(10n)}{45} = \frac{7}{15}\) \(n(10n) = 21\) \(n^210n+21=0\) 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
Joined: 14 Feb 2012
Posts: 198

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
08 May 2012, 05:37
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



Math Expert
Joined: 02 Sep 2009
Posts: 45305

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
09 May 2012, 01:33



Math Expert
Joined: 02 Sep 2009
Posts: 45305

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
07 Jun 2013, 06:11



Manager
Joined: 28 Feb 2012
Posts: 112
Concentration: Strategy, International Business
GPA: 3.9
WE: Marketing (Other)

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
08 Jun 2013, 02:08
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 nondefective 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!



Current Student
Joined: 06 Sep 2013
Posts: 1899
Concentration: Finance

Re: Defective Bulb Probability [#permalink]
Show Tags
01 Apr 2014, 06:19
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: \(\frac{n}{10}*\frac{n1}{9}=\frac{1}{15}\) > \(n(n1)=6\) > \(n=3\) 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: \(2*\frac{n}{10}*\frac{10n}{9}=\frac{7}{15}\) > \(n(10n)=21\) > \(n=3\) 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



Math Expert
Joined: 02 Sep 2009
Posts: 45305

Re: Defective Bulb Probability [#permalink]
Show Tags
01 Apr 2014, 08:27
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: \(\frac{n}{10}*\frac{n1}{9}=\frac{1}{15}\) > \(n(n1)=6\) > \(n=3\) 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: \(2*\frac{n}{10}*\frac{10n}{9}=\frac{7}{15}\) > \(n(10n)=21\) > \(n=3\) 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 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? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 01 Oct 2013
Posts: 88

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
22 Jul 2014, 09:57
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(n1) . (1) => (1/2 * n(n1))/45 =1/5 => n={3;2} => (1) sufficient . Total possibilities to either select 2 out of 10n bulbs or n defective bulbs: nC2 + (10n)C2 = 1/2 * n(n1) + 1/2 * (10n)(9n) . (2) => the probability to either select 2 out of 10n bulbs or n defective bulbs is 8/15 we have the equation: (1/2 * n(n1) + 1/2 * (10n)(9n))/45 = 8/15 => n = {3;7} => (2) sufficient



Intern
Joined: 25 Sep 2014
Posts: 7

A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
Updated on: 11 Dec 2014, 05:17
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((10n)/9) and not 2x(n/10)x((10n1)/9). Is this because they are taken simultaneously? Or is this because the second term is (9(n1))/9 = (9n+1)/9 =(10n)/9 as there is one "n" less to deduct after pulling it out in the first draw?
Appreciate your help, Many thanks
Originally posted by Parco on 11 Dec 2014, 03:49.
Last edited by Parco on 11 Dec 2014, 05:17, edited 1 time in total.



Manager
Joined: 21 Feb 2012
Posts: 59

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
11 Dec 2014, 04:49
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*(10n)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
_________________
Regards J
Do consider a Kudos if you find the post useful



Math Expert
Joined: 02 Sep 2009
Posts: 45305

Re: A box contains 10 light bulbs, fewer than half of which are [#permalink]
Show Tags
11 Dec 2014, 06:51
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((10n)/9) and not 2x(n/10)x((10n1)/9). Is this because they are taken simultaneously? Or is this because the second term is (9(n1))/9 = (9n+1)/9 =(10n)/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 n1 are defective and 9  (n1) = 10  n nondefective. So, the probability of drawing nondefective is (nondefective)/(total) = (10  n)/9.
_________________
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? Extrahard Quant Tests with Brilliant Analytics




Re: A box contains 10 light bulbs, fewer than half of which are
[#permalink]
11 Dec 2014, 06:51



Go to page
1 2
Next
[ 24 posts ]



