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

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 $= 1-\frac{12}{19}=\frac{7}{19}$