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

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A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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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.

Originally posted by marcodonzelli on 12 Jan 2008, 02:54.
Last edited by Bunuel on 05 Feb 2019, 05:52, edited 1 time in total.
Renamed the topic.
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 28 Aug 2010, 06:09
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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.
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 04 Feb 2011, 00:54
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1
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"
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 14 Jan 2008, 13:29
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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: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 27 Sep 2009, 07:06
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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: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 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
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 30 Aug 2010, 08:03
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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 = \(\frac{n}{10}*\frac{10-n}{9}+\frac{10-n}{10}*\frac{n}{9}=2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\).
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 07 Jan 2012, 15:51
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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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A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 09 Apr 2012, 07:03
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 defective.  [#permalink]

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New post 10 Apr 2012, 04:41
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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 defective.  [#permalink]

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New post 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
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 07 Jun 2013, 06:11
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 08 Jun 2013, 02:08
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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.
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 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?

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.
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 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?

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.
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 13 Dec 2015, 21:22
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


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.


When you modify the original condition and the question, it becomes defective bulbs' number:d and NOT defective bulbs' number:n. Then b+n=10 and there are 1 equation(n+d=10) and 2 variables(n,d), which should match with the number of equation. So you need 1 more equation, which is likely to make D the answer.
In 1), dC2/10C2=d(d-1)/10*9=1/15 -> d(d-1)=6 and d=3, which is unique and therefore sufficient.
In 2), dC1*(10-d)C1/10C2=7/15, d(10-d)/5*9=7/15 -> d(10-d)=21 and d=3,7. At the same time, d<5 and d=3, which is unique and therefore sufficient. So, the answer is D.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 05 Aug 2016, 07:07
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.


This little beauty should be in the 700 + category
Excellent solutions by Bunuel and fluke
One should be able to understand the answer by Bunuel Method easily.
FLUKE method is also equally elegant and one should follow it if they are finding it hard to understand the concept of symmetry in probability.

Some people are finding it hard to visualize the concept of symmetry in Probability
Here is an easy way to understand symmetry using two dice and their sum.
The minimum sum when two dice are thrown is 2 and maximum sum is 12.
We will see how the chances of getting sum from 2 to 12 are NOT EQUALLY LIKELY and how these chances are symmetrical in nature

1 way to get a sum of 2 = (1,1)
2 way to get a sum of 3 = (1,2)(2,1)
3 way to get a sum of 4 = (1,3) (2,2) (3,1)
4 way to get a sum of 5 = (1,4)(2,3)(3,2)(4,1)
5 way to get a sum of 6 = (1,5)(2,4)(3,3)(4,2)(5,1)
6 way to get a sum of 7 = (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)
5 way to get a sum of 8 = (2,6)(3,5)(4,4)(5,3)(6,2)
4 way to get a sum of 9 = (3,6)(4,5)(5,4)(6,3)
3 way to get a sum of 10 = (4,6)(5,5)(6,4)
2 way to get a sum of 11 = (5,6)(6,11)
1 way to get a sum of 12= (6,6)

As you can see the probability of getting a sum of 6 is much higher (five times higher) when two dice are thrown than of getting a sum of 2 when two dice are thrown.
Now mentally compare it with the probability of getting 6 when only dice is thrown (Probability=1/6) and getting 2 (Probability= 1/6)
This is where symmetry comes into play. IT MAKES SOME EVENTS MORE LIKELY THEN OTHER EVENTS.

Similarly the chances of Getting one good and one bad bulb in this example are symmetrical.
2 Ways to get one good bulb and one bad bulb= (First Good, Second Bad) OR (First Bad, Second Good)
Therefore while dealing with the probability of combination of one good and one bad bulb , we should keep in mind that the the probability is counted twice
Hence
2* Probability (Good Bulb and Bad bulb)= 7/15

Hope this helps !!
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Re: A box contains 10 light bulbs, fewer than half of which are defective.  [#permalink]

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New post 27 Sep 2016, 07:44
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Re: A box contains 10 light bulbs, fewer than half of which are defective.   [#permalink] 27 Sep 2016, 07:44

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