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PDF800 SET3 Question 6

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by zozo123 » Wed Aug 01, 2007 3:00 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.

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.

If someone has a strategy to quickly solve this question, thanks.
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Source: — Data Sufficiency |
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by UmanG » Wed Aug 01, 2007 3:45 am
is it A :?:
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by zozo123 » Wed Aug 01, 2007 4:31 am
The OA is D
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Re: PDF800 SET3 Question 6

by givemeanid » Wed Aug 01, 2007 5:49 am
zozo123 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 someone has a strategy to quickly solve this question, thanks.


Defective = n
Non-defective = 10-n

Two bulbs can be drawn from the box in 10C2 = 45 different ways

1. Two defective bulbs can be chosen in nC2 = n*(n-1)/2 different ways.
Probability of chosing two defective bulbs = (n*(n-1)/2)/45 = 1/15

SUFFICIENT.

2. One defective bulb can be chosen in nC1 = n different ways.
One non-defective bulb can be chosen in (10-n) different ways.
Together, one d + one non-d can be chosen in n(10-n) different ways.
Probability = n(10-n)/45 = 7/15

SUFFICIENT.


Hence, D.
So It Goes
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by jay2007 » Wed Aug 01, 2007 9:22 am
Can be solved using the formula (binomial distribution, i guess)
Let "d" be the number of defective.

Using statement#1:
Probability of getting 2 defective bulbs = ((dc2)*((10-d)c0))/(10c2)
= dc2/10c2 = 1/15
therefore d = 3.
Using statement#2:
probability of getting 1 defective and 1 good bulb
= (d/10)*((10-d)/9) + ((10-d)/10)*(d/9)
= 7/15 (given)
d can be obtained from this as 7 or 3. As d < 5, d has to be 3.

So the answer is D.
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by UmanG » Wed Aug 01, 2007 10:15 am
I missed "fewer than half"....Thanks to Jay... :)
Thanks,
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