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

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A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 12 Jan 2008, 01:54
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-box-contains-10-light-bulbs-fewer-than-half-of-which-are-99940.html
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Re: probs light bulbs [#permalink] New post 12 Jan 2008, 05:46
Either alone are sufficient. number of deffective bulbs is 3.
st1: 3/10*2/9 = 1/15
st2: 3/10*7/9 = 7/15
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Re: probs light bulbs [#permalink] New post 12 Jan 2008, 07:24
Should be D

as mentioned above, you can get n from either .

stat 1: (n/10)*(n-1/10) = 1/15. Solve for n

stat 2: (n/10)*(9-n/9) = 7/15 . Solve for n
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Re: probs light bulbs [#permalink] New post 12 Jan 2008, 10:09
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Should be D

as mentioned above, you can get n from either .

stat 1: (n/10)*(n-1/10) = 1/15. Solve for n

stat 2: (n/10)*(9-n/9) = 7/15 . Solve for n


actually it is not 9-n/9 but 10-n/9 only this works. anyway is anyone able to solve the second statement with combs????
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Re: probs light bulbs [#permalink] New post 12 Jan 2008, 14:09
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marcodonzelli wrote:
pmenon wrote:
Should be D

as mentioned above, you can get n from either .

stat 1: (n/10)*(n-1/10) = 1/15. Solve for n

stat 2: (n/10)*(9-n/9) = 7/15 . Solve for n


actually it is not 9-n/9 but 10-n/9 only this works. anyway is anyone able to solve the second statement with combs????



statement two:

(n/10)*(10-n)/9 = 7/15 simplifies to n(10-n) = 42 .. which further simplifies to n^2 - 10n +42 = 0

The above quadratic equation only has complex numbers as a solution (i.e no real solution).

i am confused. Please clarify
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Re: probs light bulbs [#permalink] New post 14 Jan 2008, 12:29
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marcodonzelli 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.


D

a fast 30-sec way without calculations:

1. the probability depends on n. for any given n there is only one value for probability. SUFF.

2. the probability is symmetric for defective and good bulbs. Therefore, there are two solutions (for example, (6,4) and (4,6)). But there is the restriction in the question: "fewer than half of which are defective". Therefore, SUFF.

a usual 2-and-more-min way: :)

1. p=n/10*(n-1)/9=1/15 ==> n²-n=6 ==> n=3,-2 ==> only positive n ==> n=3. SUFF.

2. p=n/10*(10-n)/9+(10-n)/10*n/9=7/15 ==> 10n-n²=21 ==> n²-10n+21=0 ==> n=3,7 ==> n<5 ==> n=3. SUFF.
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Re: probs light bulbs [#permalink] New post 27 Sep 2009, 06:06
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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.

Soln: n bulbs are defective and n < 5
total number of bulbs is 10.

Considering statement 1 alone
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
The probability that the two bulbs to be drawn will be defective = nC2/10C2
thus we have
= nC2/10C2 = 1/15
solving we get n = 3
Statement 1 alone is 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.
The probability that one of the bulbs to be drawn will be defective and the other will not be defective is also written as
=> nC1 * (10-n)C1/10C2 = 7/15
=> n * (10-n) = (7/15) * 10C2
=> n * (10-n) = 7 * 3
n = 3 or 7
Since n < 5, Thus n = 3 alone holds good.
Statement 2 alone is sufficient

Thus D
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Re: probs light bulbs [#permalink] New post 17 Nov 2010, 07:05
I also think this is D, but do not see the point in all the explanation, it is clear that after some calculation you will get exact number, but what is the point. Why to lose time, then it is obious that th both statement are suff.
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 03 Jan 2012, 00:41
1) nc2/10c2 = 1/2
2) [nc2 x (10-n)c1]/10c2 = 7/15
Ans. D
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 07 Jan 2012, 14:51
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Let me throw a couple of rules to crack this problem:

1. First follow the AD/BCE stragegy. That is start with the easiest statement and then use POE.
2. GMAT always prefers that both statements provide same solution, numerically. If you didn't, you effed up.

Rephrase: The only possibilities of defective:non-defective ratios are: 4:6, 3:9, 2:8, 1:9

1. Probability that both defective is 1/15. Notice that denominator is 15, out of all defective numbers in the ratios above, 3 is a multiple. Let's start with 3.
Prob = 3/10 x 2/9 = 2/30 = 1/15. So defective = 3. Suff
2. NOTE: We should try to get the same solution as 1 here and meet the S2 requirements as well. Lets start with 3 again.
Probability = 3/10 x 7/9 = 7/30. But, there are two ways this can happen, so 2 x 7/30 = 7/15. Suff.

D
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 17 Jan 2012, 22:16
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.

t = 10
n < 5

n (n-1) / t (t-1)= 1/15
15 n (n-1) = t (t-1) = 10 * 9
n (n-1) = 6
n = 3

or
ways to select 2 defective bulbs = d c2
ways to select 2 bulbs = 10c2
P(select 2 defective) = d!/2! (d-2)! * (2/ 9 * 10)
P(select 2 defective) = d!/2! (d-2)! * (1/ 9 * 5)
P(select 2 defective) = (d-1)(d) (1/2) * (1/ 45) = 1/15
(d-1)(d) = 3 * 2
d = 3

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.

2 * d/10 * (10-d)/9 = 7/15
10d - d^2 = 21
d^2 - 10d + 21 = 0
d^2 - 7d -3d + 21 = 0
d (d-7) -3 (d -7 ) = 0
d (d-7) -3 (d -7 ) = 0
(d-3) (d-7) = 0
d = 3 or 7
but defective bulbs are fewer than half -> d = 3
sufficient.
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 18 Jan 2012, 03:45
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marcodonzelli 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.

Also discussed here: a-box-contains-10-light-bulbs-fewer-than-half-of-which-are-99940.html

Hope it helps.
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 09 Apr 2012, 06:03
Hi Bunel,,
Can you please explain the second case as to why we use 2 *....( why do we take two cases)

May be with a simpler example might make things clearer!

Thanks
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 10 Apr 2012, 03:41
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shankar245 wrote:
Hi Bunel,,
Can you please explain the second case as to why we use 2 *....( why do we take two cases)

May be with a simpler example might make things clearer!

Thanks


Statement (2) says: the probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15. Now, this event can occur in two ways:
1. The first bulb is defective and the second one is not: \frac{n}{10}*\frac{10-n}{9};
2. The first bulb is NOT defective and the second one is: \frac{10-n}{10}*\frac{n}{9};

So, P=\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}.

For more check Probability chapter of Math Book: math-probability-87244.html

Hope it helps.
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 07 May 2012, 21:26
stat 1: (n/10)*(n-1/10) = 1/15. Solve for n

stat 2: (n/10)*(9-n/9) = 7/15 . Solve for n

Hence D
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 01 May 2013, 07:27
marcodonzelli 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.

here the bulbs are drawn simultaneously so the better approach would be : statement 1: nc2/10c2 =1/15 ;solve for n and choose a positive value.
statement 2: (nc1*(10-n)c1)/10c2 =7/15. solve for n and choose a value less than 5 (because fewer than half are defective)
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 02 May 2013, 02:45
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 03 May 2013, 20:26
Hi, For me the answer for this DS is E ...... but different guys gave different answers for this, Can Elaborate & discuss, this Pls.
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 04 May 2013, 01:13
FRANCISCODake wrote:
Hi, For me the answer for this DS is E ...... but different guys gave different answers for this, Can Elaborate & discuss, this Pls.


the option is D and not E.because each of the statements is sufficient to find the number of faulty bulbs..
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Re: A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 04 May 2013, 03:18
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FRANCISCODake wrote:
Hi, For me the answer for this DS is E ...... but different guys gave different answers for this, Can Elaborate & discuss, this Pls.


The correct answer is D, not E.

In case of any questions please post here: a-box-contains-10-light-bulbs-fewer-than-half-of-which-are-99940.html
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Re: A box contains 10 light bulbs, fewer than half of which are   [#permalink] 04 May 2013, 03:18
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