One ticket is selected at random from $$100$$  tickets numbered $$00,\,01,\,02,\, ....,\,98,\,99.$$      If $${x_1}$$ and $${x_2}$$ denotes the sum and product of the digits on the tickets, then $$P\left( {{x_1} = \frac{9}{{{x_2}}} = 0} \right)$$    is equal to :

Solution :
$$\eqalign{ & {\text{Let the number selected by }}xy{\text{. Then,}} \cr & x + y = 9,\,0 \leqslant x,\,y \leqslant 9{\text{ and}} \cr & xy = 0 \Rightarrow x = 0,\,y = 9 \cr & {\text{or }}y = 0,\,x = 9 \cr & P\left( {{x_1} = \frac{9}{{{x_2}}} = 0} \right) = \frac{{P\left( {{x_1} = 9 \cap {x_2} = 0} \right)}}{{P\left( {{x_2} = 0} \right)}} \cr & {\text{Now, }}P\left( {{x_2} = 0} \right) = \frac{{19}}{{100}}{\text{ and}} \cr & P\left( {{x_1} = 9 \cap {x_2} = 0} \right) = \frac{2}{{100}} \cr & \Rightarrow P\left( {{x_1} = \frac{9}{{{x_2}}} = 0} \right) = \frac{{\frac{2}{{100}}}}{{\frac{{19}}{{100}}}} = \frac{2}{{19}} \cr} $$