$$10$$ apples are distributed at random among $$6$$ persons. The probability that at least one of them will receive none is :

Solution :
The required probability $$ = 1 - $$   probability of each receiving at least one
$$ = 1 - \frac{{n\left( E \right)}}{{n\left( S \right)}}.$$
Now, the number of integral solutions of $${x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = 10$$        such that $${x_1} \geqslant 1,\,{x_2} \geqslant 1,\,......,\,{x_6} \geqslant 1$$       gives $$n\left( E \right)$$  and the number of integral solutions of $${x_1} + {x_2} + ...... + {x_5} + {x_6} = 10$$       such that $${x_1} \geqslant 0,\,{x_2} \geqslant 0,\,......,\,{x_6} \geqslant 0$$       gives $$n\left( S \right).$$
$$\therefore $$  the required probability $$ = 1 - \frac{{{}^{10 - 1}{C_{6 - 1}}}}{{{}^{10 + 6 - 1}{C_{6 - 1}}}} = 1 - \frac{{{}^9{C_5}}}{{{}^{15}{C_5}}} = \frac{{137}}{{143}}.$$