Answer: Option A
$$P(X) = \frac{1}{5},$$   $$P(Y) = \frac{1}{4},$$   $$P(Z) = \frac{1}{3}$$
Required probability:
$$ = \left[ {P\left( A \right)P\left( B \right)\left\{ {1 - P\left( C \right)} \right\}} \right]$$     $$ + $$ $$\left[ {\left\{ {1 - P\left( A \right)} \right\}P\left( B \right)P\left( C \right)} \right] + $$     $$\left[ {P\left( A \right)P\left( C \right)\left\{ {1 - P\left( B \right)} \right\}} \right] + $$     $$\left[ {P\left( A \right)P\left( B \right)P\left( C \right)} \right]$$
$$ = \left[ {\frac{1}{4} \times \frac{1}{3} \times \frac{4}{5}} \right] + $$   $$\left[ {\frac{3}{4} \times \frac{1}{3} \times \frac{1}{5}} \right] + $$   $$\left[ {\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}} \right] + $$   $$\left[ {\frac{1}{4} \times \frac{1}{3} \times \frac{1}{5}} \right]$$
$$\eqalign{
& = \frac{4}{{60}} + \frac{3}{{60}} + \frac{2}{{60}} + \frac{1}{{60}} \cr
& = \frac{{10}}{{60}} \cr
& = \frac{1}{6} \cr} $$