Probability

Please remember that Maximum portability is 1.

So we can get total probability of non-defective bulbs and subtract it from 1 to get total probability of defective bulbs.

Total cases of non defective bulbs = ¹⁶C₂ = 16 × 152 × 1 = 120

= ²⁰C₂ = $\genfrac{}{}{1px}{}{\mathrm{20\; \times \; 19}}{\mathrm{2\; \times \; 1}}$ = 190

= $\genfrac{}{}{1px}{}{120}{\mathrm{190}}$ = $\genfrac{}{}{1px}{}{12}{\mathrm{19}}$

Probability of at least one defective = 1 − 1$\genfrac{}{}{1px}{}{12}{\mathrm{19}}$ = $\genfrac{}{}{1px}{}{7}{\mathrm{19}}$