$\frac{7}{19}$

$\frac{6}{19}$

$\frac{5}{19}$

$\frac{4}{19}$

A. $\frac{7}{19}$

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 $=1-\frac{12}{19}=\frac{7}{19}$