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

Correct Answer : A.

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
${}^{16}c_2=\frac{16\times 15}{2\times 1}=120$
Total cases $={}^{20}c_2=\frac{20\times 19}{2\times 1}=190$
probability $= \frac{120}{190}=\frac{12}{19}$
P of at least one defective $= 1-\frac{12}{19}=\frac{7}{19}$