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:

$P(A)=.4$, $P(B)=.3$, $P(C)=.25$, $P(A\cap B)=.23$, $P(A\cup B\cup C)=.49$, $P(A \mid C)=.88$, $P(C \mid B)=.7$

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.