Solution 4.9.9.1.

\begin{equation*} P(A \cap B \cap C) = \frac{2}{5} \frac{3}{4} \frac{1}{6}. \end{equation*}
By the corollary for independent events, complements also maintain a similar independence. So
\begin{equation*} P(A^c \cap B^c \cap C) = \frac{3}{5} \frac{1}{4} \frac{1}{6} . \end{equation*}
To complete the third part, use the inclusion/exclusion resultĀ 4.3.9 for dealing with three sets.
in-context