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# A box contains 10 light bulbs, fewer than half of which are

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A box contains 10 light bulbs, fewer than half of which are [#permalink]

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09 Nov 2008, 18:25
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

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09 Nov 2008, 22:40
gorden 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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

my intuition tells me that the ans is D....

1) x/10 * (x-1)/9 = 1/15
2) x/10 * (9-x)/9 = 7/15

both can get x=?

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09 Nov 2008, 23:13
Sion wrote:
gorden 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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

my intuition tells me that the ans is D....

1) x/10 * (x-1)/9 = 1/15
2) x/10 * (9-x)/9 = 7/15
both can get x=?

I got A.

How did you get (9-x)/9 in the second stmt? To me, it should be (10-x)/9 and with this, there is no real value for x. Hence, A.

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10 Nov 2008, 00:26
scthakur wrote:
Sion wrote:
gorden 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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

my intuition tells me that the ans is D....

1) x/10 * (x-1)/9 = 1/15
2) x/10 * (9-x)/9 = 7/15
both can get x=?

I got A.

How did you get (9-x)/9 in the second stmt? To me, it should be (10-x)/9 and with this, there is no real value for x. Hence, A.

oops.. a little mistake
but although i agree that stmt2 is not sufficient, it should be proved like this:

if there are 10 bulbs, x is defective, then 10-x is not defective
in stmt2, one of the bulbs to be drawn will be defective, so the probability is x/10, now the number of remainder bulb is 10-1, and the number of remainder bulbs which is not defective is 10-(x-1).
hence, it should be x/10 * [10-(x-1)]/9 = 7/15

x=14, but there are only 10 bulbs, thus stmt2 is not suff.

open to discussion

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10 Nov 2008, 00:31
B is insufficient.

if n is defective, then (10-n) is not defective.

For stmt2: either the first one is defective or the second one is defective.

Hence, n/10*(10-n)/9 + (10-n)/10*n/9 = 7/15
or, n = 3, 7.

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10 Nov 2008, 00:44
scthakur wrote:
B is insufficient.

if n is defective, then (10-n) is not defective.

For stmt2: either the first one is defective or the second one is defective.

Hence, n/10*(10-n)/9 + (10-n)/10*n/9 = 7/15
or, n = 3, 7.

you are right, i give up~

what a nice question!

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10 Nov 2008, 01:13
the stem says fewer than half are defective...so n<5.
from both (1) and (2) we get n=3.

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10 Nov 2008, 02:32
prasun84 wrote:
the stem says fewer than half are defective...so n<5.
from both (1) and (2) we get n=3.

What a catch! Awesome! Yes, answer should be D.

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10 Nov 2008, 03:42
infact...plugging in values of 2,3,or 4 can also be used to solve...

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Re: bulbs--28   [#permalink] 10 Nov 2008, 03:42
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