Throughout these calculations, you can presume that the first five numbers are selected independently from the Mega Ball number. However, the first five numbers are selected without replacement so computing probabilities with those does not allow for independence. This part is hypergeometric with the \(n_1 = 5\) numbers you selected being the "desired" numbers and the Lottery Commission picking a subset of size r = 5 from the 75 possible numbers. So, your likelihood of matching all five would be
\begin{equation*}
\frac{\binom{5}{5} \cdot \binom{70}{0}}{\binom{75}{5}} = \frac{1}{17259390}.
\end{equation*}
Multiplying this by the 1 chance in 15 that you also match the Mega Ball gives
\begin{equation*}
P(\text{Match 5 plus Mega Ball}) = \frac{1}{17259390} \cdot \frac{1}{15} = \frac{1}{258,890,850}.
\end{equation*}
To match only 5 means you also MUST miss the Mega Ball which has probability 14/15 to give
\begin{equation*}
\frac{1}{17259390} \cdot \frac{14}{15} = \frac{1}{17259390 \cdot \frac{15}{14}} \approx \frac{1}{18492204}.
\end{equation*}
Continue in this manner to determine the other odds.