How would I solve this problem if the conditional probability for $\mathsf P(B \mid A)$ is not listed? Also, in Bayes' Theorem only intersection is listed, not Union. How would I find Union?
$\mathsf P( A \cup B)$ if it is given that $\mathsf P(A)=1/3$ and $\mathsf P(B \mid A^c)=1/4$
Use the following equalities: $\begin{cases}\begin{align} \mathsf P(A\cup B) &= \mathsf P(A)+\mathsf P(B) - \mathsf P(A\cap B) & \tag{Inclusion/Exclusion} \\[2ex] \mathsf P(A\cap B) &= \mathsf P(B)-\mathsf P(A^c\cap B) & \tag{Complementation} \\ \mathsf P(A) & = 1 - \mathsf P(A^c) \\[2ex] \mathsf P(B\mid A^c) &= \frac{\mathsf P(A^c\cap B)}{\mathsf P(A^c)} & \tag{Bayes' Rule} \end{align}\end{cases}$