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DS – Probability-Defective Bulbs

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Manager
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DS – Probability-Defective Bulbs [#permalink] New post 16 Apr 2007, 08:51
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A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Guys, could you please help with the problem below:

28-6. 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.
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 [#permalink] New post 16 Apr 2007, 10:47
I think (1) works, but I don't think (2) is sufficient, so I would go with A.

Does anyone have the answer?
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 [#permalink] New post 16 Apr 2007, 11:41
Answer= D (both statements are sufficient)

Here's how

State I)Lets say 'N' bulbs are defective. Number of ways both bulbs are defective
= NC2
Total Number of ways =10C2

probablity is NC2/10C2 = 7/15, Solving this we will have N=7 or 3, Since problem said N id less than half (half of 10). Hence N=3


Statement II) Atleast one defective is NC1+(10-N)C1 ways. Since probablity is given. we can calculate n in the same way as above.

So both Statements are enough.
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 [#permalink] New post 16 Apr 2007, 16:40
Yes, it's D as vijay2001 explained. :)
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 [#permalink] New post 16 Apr 2007, 17:08
# of bulbs defective can be 1-4 (fewer than half)

St1:
We know n/10 * (n-1)/9 = 1/15
n(n-1) = 6
n^2 - n - 6 = 0

n = 3 or n = -2 (invalid)

Sufficient.

St2:
so we're told [n/10 * (10-n)/9] + [(10-n)/10 * n/9] = 7/15

(n(10-n)/90)*2 = 7/15
n(10-n)/90 = 7/30
n(10-n) = 21
10n - n^2 = 21
n^2 - 10n + 21 = 0
n = 3 or n = 7 (invalid)

Sufficient.

Ans D
  [#permalink] 16 Apr 2007, 17:08
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DS – Probability-Defective Bulbs

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