Striver 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 the 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.