For our probability homework we need to calculate $ p(\divideontimes) $ and $ p(\neg \divideontimes) $ (Which I assume is $p(\neg \divideontimes) = 1- p(\divideontimes) $). I am however stuck on how to proceed to do this. I assumed that I needed to use Bayes' Theorem ($ p(x \mid y) = \frac{p(y \mid x)p(x)}{p(y)}$) to calculate $ p(\divideontimes) $ by filling in part of the equation so I can reduce $ p(\divideontimes)$ from the formula, however I am stuck while trying to do this.
I don't really know how to proceed on this question. A push in the right direction would be greatly appreciated.
I believe you need to use the Law of Total Probability (which is used in Bayes' theorem, but Bayes' theorem isn't needed here).
The law states that if a space $\mathscr{F}$ is partitioned into $n$ mutually exclusive (disjoint) and exhaustive events $B_i$ for $i=1,2,3, \ldots, n$ such that $\mathbb{P}(B_i) > 0$ and if $A \subset \mathscr{F}$, then
\begin{align} \mathbb{P}(A)&=\sum_{i=1}^{n} \ \mathbb{P}(A \cap B_i) \\ &= \sum_{i=1}^{n} \ \mathbb{P}(A |B_i) \ \mathbb{P}(B_i) \end{align}
Think you can handle it from here? Let me know if something is unclear.