A box contains 20 electric bulbs, out of which 4 are defective. Two balls are chosen at random from this box. The probability that at least one of them is defective, is -

A. $$\frac{4}{19}$$

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

C. $$\frac{12}{19}$$

D. $$\frac{21}{95}$$

E. None of these

Answer: Option B

Solution(By Examveda Team)

P (none is defective)
= n (E) =  $$\frac{{{}^{16}\mathop C\nolimits_2 }}{{{}^{20}\mathop C\nolimits_2 }} = $$ $$\left( {\frac{{16 \times 15}}{{2 \times 1}} \times \frac{{2 \times 1}}{{20 \times 19}}} \right)$$     $$ = \frac{{12}}{{19}}$$
P (at least 1 is defective)
=$$\left( {1 - \frac{{12}}{{19}}} \right)$$   $$ = \frac{7}{{19}}$$