A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective is	Select the correct option.
A) 7 / 19
B) 6 / 19
C) 5 / 19
D) 4 / 19
E) none of these

Rationale: "explanation : please remember that maximum portability is 1 . so we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs . so here we go , total cases of non defective bulbs 16 c 2 = 16 ∗ 15 / 2 ∗ 1 = 120 total cases = 20 c 2 = 20 ∗ 19 / 2 ∗ 1 = 190 probability = 120 / 190 = 12 / 19 p of at least one defective = 1 − 12 / 19 = 7 / 19 option a"
Answer: A