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

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Joined: 16 Aug 2015
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Re: A box contains 10 light bulbs, fewer than half of which are  [#permalink]

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13 Dec 2015, 20:22
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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A box contains 10 light bulbs, fewer than half of which are  [#permalink]

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05 Aug 2016, 06:07
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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A box contains 10 light bulbs, fewer than half of which are  [#permalink]

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27 Sep 2016, 06:44
Here is my two cents
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Joined: 11 Sep 2013
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Concentration: Finance, Finance
Re: A box contains 10 light bulbs, fewer than half of which are  [#permalink]

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22 Dec 2017, 06:00
1
I don't know why is every one even trying to calculate anything here.

Both the statements clearly mention about probability. No statement says something like this" Probability of defective of not defective or anything related to this question is less than or more than some numbers"

Since both statements are giving two equations, not inequalities, we can somehow find the number of defective bulb.
So Each statement is sufficient
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Re: A box contains 10 light bulbs, fewer than half of which are  [#permalink]

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25 Dec 2018, 10:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A box contains 10 light bulbs, fewer than half of which are &nbs [#permalink] 25 Dec 2018, 10:05

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