for the room to be lit, it is sufficient that atleast 1 bulb is lit (not necessary all 3 should work)

So find, 1 - P(all 3 are defective)

1 - 4C3/12C3 = 54/55

The concept u need to understand is that for the room to be lit, having only one working bulb is also sufficient. Its not necessary that all three bulbs have to be working. Let W stand for working and D stand for defective. So we could have (DWW)3!/2!+(DDW)3!/2!+(WWW). What this translates into is, If three bulbs are picked up at random, what is the probability that atleast one works. Since the above calculation is a bit time consuming we can find the complementary i.e. the probability that no bulb works out of the three picked up.

# of possible outcomes=12C3=220
Number of ways to pick only 3 defective bulbs=4C3=4
Probability=4/220=1/55

Now, 1-1/55=54/55 Hence C