I have a question about the following:

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.

Because the question says "simultaneously", does this mean that you wouldn't calcultate (1) by saying 3/10 * 2/9 =1/15 ?

Also, how would you calculate number (2)?

Thanks!

Is the answer D?
If so, I computed the defective bulbs to be 3.
From clue1: 1/15 ==> 6/90 ==> 3*2/10*9 Given defective bulbs are fewer than 5, only poss is 3*2(3-D,7-G)
From clue2:7/15==> 42/90 ==> (7*3)/10*9+(3*7)/10*9=42/90 Given defective bulbs are fewer than 5, betweeen 7 and 3, you can assign 3 for defective bulbs

D is correct. Thanks for the explanation!