Data Sufficiency

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.

OA is D.

Here,the question asks that the two bulbs be drawn simultaneously(bulbs are not drawn one after another).So how
do we calculate this type of probability?

Don't worry about the simultaneous stuff, just solve it as normal. You can eye these, but I wrote out the long way to solve, since that seems to be your main question

A- Sufficient
(x/10)*((x-1)/(10-1))
(x^2-x)/90
x^2-x-6=0
x= 3 or -2
so x= 3

B- Sufficient
Slightly trickier
(x/10)*((10-x)/(10-1)) + ((10-x)/10)*(x/(10-1)) = 7/15
((10x-x^2)/90) + ((10x-x^2)/90) = 7/15
(-2x^2+20x)/90 = 7/15
-2x^2+20x-41=0
-2(x^2-10x+21) = 0
-2(x-7)(x+3) = 0

So x could be 3 or 7. Since we know that it is less than 5 the answer is 3