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light bulbs-probability

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light bulbs-probability

by dikku07 » Thu Sep 24, 2009 10:51 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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light bulbs-probability

by shankykm » Thu Sep 24, 2009 12:15 pm
Is the answer A?

As from first statement we get no.of defective bulbs to be 3, which is sufficient.

But from second statement we get two values for n(no.of defective bulbs)3 and 7. Hence we cannot be so sure as which one is right. SO statement two is not sufficient

Hence answer is A
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by grockit_jake » Thu Sep 24, 2009 3:27 pm
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.

If we were given n, this would be straight forward, but we can use the variable in the same way we would use a number. For (1).

[n/10]*(n-1)/9 = 1/15
n(n-1) = 6
n^2 - n - 6 = 0
(n-3)(n+2) = 0
n=3, n=-2. Since it's impossible for n<0, n=3.

For (2):

[n/10]*[10-n)/ 9]*2 = 7/15
n(10-n) = 42
10n - n^2 = 21
n^2 - 10n + 21 = 0
(n-3)(n-7)=0
n=3,n=7.

A it is.
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by woo » Fri Sep 25, 2009 2:28 am
grockit_jake wrote: For (2):

[n/10]*[10-n)/ 9]*2 = 7/15
n(10-n) = 42
10n - n^2 = 21
n^2 - 10n + 21 = 0
(n-3)(n-7)=0
n=3,n=7.

A it is.
In the first line, why do you have to multiply by 2?
Is it because we have to consider 2 cases: one with defective chosen first and the other with normal bulb chosen first?
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by mohitsharda » Fri Sep 25, 2009 4:01 am
grockit_jake 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.

If we were given n, this would be straight forward, but we can use the variable in the same way we would use a number. For (1).

[n/10]*(n-1)/9 = 1/15
n(n-1) = 6
n^2 - n - 6 = 0
(n-3)(n+2) = 0
n=3, n=-2. Since it's impossible for n<0, n=3.

For (2):

[n/10]*[10-n)/ 9]*2 = 7/15
n(10-n) = 42
10n - n^2 = 21
n^2 - 10n + 21 = 0
(n-3)(n-7)=0
n=3,n=7.

A it is.
The question states that fewer than half are defective
so, even though we get n =3,7 from the second statement,
we can only have n =3.
So it is also sufficient
Hence answer is D
MS
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by dikku07 » Sat Sep 26, 2009 6:35 am
OA is D

But could you pls explain the 2nd statement- how did you guys solve it?
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by rsruparelia » Sat Sep 26, 2009 8:56 am
Can you please explain, why did you multiply the 2nd statement with 2??
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by grockit_jake » Sat Sep 26, 2009 2:53 pm
Yes, you are right. I missed the "less than half" part. In that case, both 1 and 2 are sufficient.

I multiplied by 2 because order mattered here: [n/10]*[10-n)/ 9].

Namely, the first time necessitates defective and the 2nd necessitates non-defective. However, in the question, order does not matter, so essentially it's:

[n/10]*[10-n)/ 9] or [n/10]*[10-n)/ 9] =
[n/10]*[10-n)/ 9] + [n/10]*[10-n)/ 9] =
2*[n/10]*[10-n)/ 9]
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by grockit_jake » Sat Sep 26, 2009 2:54 pm
Or you can see that since we are multiplying fractions, and adding 2 sets that are identical but for order, then you can just double the original.
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