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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1. total way of drawing 2 bulbs are 10C2 =45bacali 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.
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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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
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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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....
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....
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