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# DS â€“ Probability-Defective Bulbs

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16 Apr 2007, 09:51
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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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16 Apr 2007, 11:47
I think (1) works, but I don't think (2) is sufficient, so I would go with A.

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16 Apr 2007, 12:41
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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16 Apr 2007, 17:40
Yes, it's D as vijay2001 explained.
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16 Apr 2007, 18:08
# 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
16 Apr 2007, 18:08
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