Explanation

(c)
Probability for selecting a white ball $= \frac{2}{6} = \frac{1}{3}$
Probability for selecting a black ball $= \frac{4}{6} = \frac{2}{3}$
$\therefore$ Required probability $= {\ ^{5}C}_{5}\left( \frac{1}{3} \right)^{5}\left( \frac{2}{3} \right)^{0} + {\ ^{5}C}_{4}\left( \frac{1}{3} \right)^{4}\left( \frac{2}{3} \right)$
$= \left( \frac{1}{3} \right)^{5} + 5\left( \frac{1}{3} \right)^{4}\left( \frac{2}{3} \right)$
$= \left( \frac{1}{3} \right)^{4}\left\lbrack \frac{1}{3} + 5 \bullet \frac{2}{3} \right\rbrack$
$= \frac{11}{3^{5}} = \frac{11}{243}$