[tex] \Large \mathcal{SOLUTION:} [/tex]
[tex] \begin{array}{l} \textsf{By Principle of Inclusion and Exclusion,} \\ \\ \qquad\:\: P(E) = 1 - P(E') \\ \\ \begin{aligned} P(\textsf{atleast 1 black ball}) &= 1 - P(\textsf{No black ball}) \\ \\ P(\textsf{atleast 1 black ball}) &= 1 - \dfrac{{}^6C_3}{{}^9C_3} \\ \\ P(\textsf{atleast 1 black ball}) &= 1 - \dfrac{\frac{6!}{3! \: 3!}}{\frac{9!}{3! \: 6!}} \\ \\ P(\textsf{atleast 1 black ball}) &= 1 - \dfrac{\frac{6\cdot 5\cdot 4}{6}}{\frac{9\cdot 8\cdot 7}{6}} \\ \\ P(\textsf{atleast 1 black ball}) &= 1 - \dfrac{20}{84} \\ \\ P(\textsf{atleast 1 black ball}) &= \dfrac{84 - 20}{84} \\ \\ P(\textsf{atleast 1 black ball}) &= \dfrac{64}{84} \\ \\ P(\textsf{atleast 1 black ball}) &= \boxed{\frac{16}{21}} \end{aligned} \end{array} [/tex]
