Need explanation for probability problem (URGENT)

[akpareek]

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 2-bulbs to be drawn will be defective is 1/15.
2) The probability that the 2-bulbs to be drawn will be defective and the other will not be defective is 7/15.

[sanjoy18]

2 statement is wrong..it should be one is defective and another is not
n defective bulbs and where n< 5

St(1)The probability that the two bulbs to be drawn will be defective is 1/15.
The probability that the two bulbs to be drawn will be defective = nC2/10C2
thus we have
= nC2/10C2 = 1/15
solving we get n = 3
Statement 1 alone is sufficient

st(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.
hence
nC1 * (10-n)C1/10C2 = 7/15
>> n * (10-n) = (7/15) * 10C2
>> n * (10-n) = 7 * 3
n = 3 or 7
Since n < 5, Thus n = 3 alone holds good.
Statement 2 alone is sufficient

hence D

[[email protected]]

Hi All,

sanjoy18 correctly caught the error and explained the "math" behind this question (nicely done!).

I wanted to emphasize a particular point about GMAT Quant questions that is worth noting. GMAT writers will give you information that matters (in essence, there's no filler), so when this prompt mentions that fewer than half of the lightbulbs are defective (meaning n < 5), you should PAY ATTENTION to that fact. The real GMAT killers will ask themselves "WHY does THAT matter? What role will that play later on? Will it LIMIT the number of possibilities?"

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

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.

Statement 1 gives something away: since P(both bulbs are defective) > 0, there must be at least 2 defective bulbs.
Since the question stems indicates that n<5, we know that the only possible values are n=2, n=3, or n=4.

Statement 1: P(both are defective) = 1/15.
If n=2, P(both are defective) = 2/10 * 1/9 = 1/45. Doesn't work.
If n=3, P(both are defective) = 3/10 * 2/9 = 1/15. This works.
Clearly, n=4 can't work, since it will increase all the numerators.
Thus, n=3.
Sufficient.

Statement 2: P(exactly 1 is defective) = 7/15.
Since only n=3 worked in statement 1, we should start with n=3.

If n=3, P(1st is defective and the 2nd is not) = 3/10 * 7/9 = 7/30.
Since P(1st is not defective and the 2nd is defective) will yield the same probability, the result above must be multiplied by 2:
P(exactly 1 is defective) = 2 * 7/30 = 7/15. This works.

Clearly n=4 can't work, since it will increase all the numerators.
Using similar logic, n=2 can't work, since it will decrease all the numerators.
Thus, n=3.
Sufficient.

The correct answer is D.