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by bacali » Fri Nov 14, 2008 11:59 am
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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Source: — Data Sufficiency |
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by EricLien9122 » Fri Nov 14, 2008 1:03 pm
In my opinion, the answer is D.

I will go in to details if my answer is correct.
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by gmat009 » Fri Nov 14, 2008 1:07 pm
Is it D
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by bacali » Fri Nov 14, 2008 1:44 pm
OA: D
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by bacali » Fri Nov 14, 2008 1:45 pm
EricLien9122 wrote:In my opinion, the answer is D.

I will go in to details if my answer is correct.
please explain
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Re: Another Question

by sushanta57021 » Fri Nov 14, 2008 3:05 pm
bacali 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.
1. total way of drawing 2 bulbs are 10C2 =45
The probability that the two bulbs to be drawn will be defective is

nC2 = n(n-1)/2

so probability= n(n-1)/2x45=1/15
n=3

Sufficient.

2. The probability that one of the bulbs to be drawn will be defective and the other will not be defective is

nc1x(10-n)C1=n(10-n)

so, n(10-n)/45= 7/15

n=3,n=7

as n<5
n=3

sufficient.

hence D
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by Fab » Sat Nov 15, 2008 2:17 pm
Hi,

I am having some trouble with statement 2...

My logic is this:

Statement 1:

Number of defective bulbs are between 1 and 4, so:

X/10 * Y/9 = 1/15
Solving: X*Y=90/15
X*Y=6 ----->So X must be 3 and Y 2
SUFF

Statement 2:

X/10* Y/9 = 7/15
Solving X*Y=42
But the smallest values I get are 7 for X and 6 for Y...

wrong...?

THANKS
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by jimmiejaz » Mon Nov 17, 2008 4:08 am
Hi,

7/15 also includes the probability when first the non defective bulb is taken out and then the defective bulb is taken out.
What i mean here is that there are 2 ways to take out one defective bulb and one non defective way. ie. D,ND or ND,D
so, you have to divide it by half.
then you will get xy=(42/2) = 21
which will give you 7,3.

Hope this solves ur query....
What if i have not yet beat the beast, I know i will beat it!!!!!!!!
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