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

 It is currently 21 Oct 2018, 18:29

### 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

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

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Apr 2010
Posts: 117
A box contains 10 light bulbs, fewer than half of which are  [#permalink]

### Show Tags

28 Aug 2010, 02:54
3
18
00:00

Difficulty:

45% (medium)

Question Stats:

71% (02:10) correct 29% (02:22) wrong based on 865 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.
Math Expert
Joined: 02 Sep 2009
Posts: 50009

### Show Tags

28 Aug 2010, 06:09
7
20
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.

Hope it's clear.
_________________
##### General Discussion
VP
Joined: 17 Feb 2010
Posts: 1158

### 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'

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

Thanks
Intern
Joined: 05 Nov 2009
Posts: 29

### Show Tags

29 Aug 2010, 20:00
1
Seekmba,

There are 2 ways to draw 2 bulbs in which 1 works and 1 is defective.
VP
Joined: 17 Feb 2010
Posts: 1158

### Show Tags

29 Aug 2010, 21:07
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.
Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 629
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas

### Show Tags

29 Aug 2010, 21:54
1
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

Math Expert
Joined: 02 Sep 2009
Posts: 50009
A box contains 10 light bulbs, fewer than half of which are  [#permalink]

### Show Tags

30 Aug 2010, 08:03
7
3
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 = $$\frac{n}{10}*\frac{10-n}{9}+\frac{10-n}{10}*\frac{n}{9}=2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}$$.
_________________
Manager
Joined: 17 Mar 2010
Posts: 145

### Show Tags

31 Aug 2010, 04:32
Bunuel Rocks.....
Intern
Joined: 03 Feb 2011
Posts: 2

### 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.
Retired Moderator
Joined: 20 Dec 2010
Posts: 1835

### Show Tags

04 Feb 2011, 00:54
3
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

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

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

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

n=3 and n=7

Since,
0<=n<5
n = 3

Sufficient.

Ans: "D"
_________________
Manager
Joined: 14 Feb 2012
Posts: 136
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: 50009
Re: A box contains 10 light bulbs, fewer than half of which are  [#permalink]

### Show Tags

09 May 2012, 01:33
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

Hope it helps.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 50009
Re: A box contains 10 light bulbs, fewer than half of which are  [#permalink]

### Show Tags

07 Jun 2013, 06:11
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

_________________
Manager
Joined: 28 Feb 2012
Posts: 112
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
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
Joined: 06 Sep 2013
Posts: 1764
Concentration: Finance

### 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: $$\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}$$ --> $$n(n-1)=6$$ --> $$n=3$$ 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.

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: 50009

### 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: $$\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}$$ --> $$n(n-1)=6$$ --> $$n=3$$ 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.

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.
_________________
Manager
Joined: 01 Oct 2013
Posts: 76
Schools: Haas '16, AGSM '16
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(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
Joined: 25 Sep 2014
Posts: 5
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((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?

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: 58
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*(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.

_________________

Regards
J

Do consider a Kudos if you find the post useful

Math Expert
Joined: 02 Sep 2009
Posts: 50009
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((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?

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

Go to page    1   2    Next  [ 24 posts ]

Display posts from previous: Sort by