Consider the following information about travelers on vacation: $40$% check work email, $30$% use a cell phone, $25$% bring a laptop with them, $23$% both check work email and use a cell phone, and $51$% neither check work email nor use a cell phone nor bring a laptop. In addition $88/100$ who bring a laptop also check work email, and $70/100$ who use a cell phone also bring a laptop.
What is the probability that someone who brings a laptop on vacation also uses a cell phone?
I let $A$ represent check work email, $B$ represent use cell phone, and $C$ represent brings laptop. I have the following probabilities:
$$\begin{align}P(A)&=.4, \\P(B)&=.3, \\P(C)&=.25,\\ P(A\cap B)&=.23, \\P(A\cup B\cup C)&=.49,\\ P(A \mid C)&=.88, \\P(C \mid B)&=.7\end{align}$$
I found a solution online saying to use Bayes' Theorem to solve this part of the problem but I do not understand why to use Bayes' Theorem. Here is a picture to the solution I am referring to:
Bayes' Theorem does not look like what the solution says to use. In my textbook, the theorem looks like this:
Did the person that wrote the solution simplify something? It does not look like either of these two forms.
Also, for problems like these, is there a general rule on when to use Bayes' Theorem and the rule for Total Probability? I cannot figure out when to use what.