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

Given: \(bulbs=10\) and \(defective=n<5\). Question: \(n=?\)

(1) The probability that the two bulbs to be drawn will be defective is 1/15 --> clearly sufficient, as probability, \(p\), of drawing 2 defective bulbs out of total 10 bulbs, obviously depends on # of defective bulbs, \(n\), so we can calculate uniques value of \(n\) if we are given \(p\).

To show how it can be done: \(\frac{n}{10}*\frac{n-1}{9}=\frac{1}{15}\) --> \(n(n-1)=6\) --> \(n=3\) or \(n=-2\) (not a valid solution as \(n\) represents # of defective bulbs and can not be negative). Sufficient.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15 --> also sufficient, but a little bit trickier: if it were 3 defective and 7 good bulbs OR 7 defective and 3 good bulbs, then the probability of drawing one defective and one good bulb would be the same for both cases (symmetric distribution), so info about the probability, 7/15, of drawing one defective and one good bulb would give us 2 values of \(n\) one less than 5 and another more than 5 (their sum would be 10), but as we are given that \(n<5\), we can stiil get unique value of \(n\) which is less than 5.

To show how it can be done: \(2*\frac{n}{10}*\frac{10-n}{9}=\frac{7}{15}\) --> \(n(10-n)=21\) --> \(n=3\) or \(n=7\) (not a valid solution as \(n<5\)). Sufficient.

Answer: D.

Hope it's clear.