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

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Joined: 13 Jun 2004
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A box contains 10 light bulbs, fewer than half of which are [#permalink] New post 18 Oct 2004, 18:05
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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.

Please explain the steps.
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 [#permalink] New post 18 Oct 2004, 19:38
Ans D

From Statment 1 we can obtain the following equation

(n/10)((n-1)/9) = 1/15
= (n^2-n)/90 = 1/15
= n^2 - n = 6
= n^2 - n - 6 = 0
=(n-3)(n+2) = 0
So n = 3 or n = -2 And since we know that n can't be negative
n = 3

From Statement 2 we can obtain the folllowing equation

2*((n/10)((10-(n-1))/9) = 7/15
= 20n-2n^2+2n = 42
= 2n^2-22n+42 = 0
=(2n-14)(n-3) = 0
So n = 7 or n= 3 And since we know that n is less than half,
n = 3
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 [#permalink] New post 18 Oct 2004, 23:55
OA is D :done

Thanks Nzgmat, your explanation is clear and it was exactly what I was waiting for because I got the OA but no explanations :?
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 [#permalink] New post 26 Oct 2004, 21:03
Hey nzgmat, can you explain this
Quote:
From Statement 2 we can obtain the folllowing equation

2*((n/10)((10-(n-1))/9) = 7/15
= 20n-2n^2+2n = 42
= 2n^2-22n+42 = 0
=(2n-14)(n-3) = 0
So n = 7 or n= 3 And since we know that n is less than half,
n = 3



My understanding is that it should be :

2*((n/10)((9-(n-1))/9) = 7/15

so the equation becomes :

n(10-n) =21 , so n = 7 or 3, and since n < 5 , it should be 3.


Alternatively we can think of the equation as :

nC1 * (10-n)C1 / 10C2 = 7/15
  [#permalink] 26 Oct 2004, 21:03
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