Question:
The table below shows the success rates and numbers of treatments for treatments involving both small and large kidney stones, where Treatment A includes open surgical procedures and Treatment B includes closed surgical procedures. The numbers in parentheses indicate the number of success cases over the total size of the group.
$$\begin{array}{|c|c|c|c|} \hline \text{Stone size\Treatment} & \text{Treatment A}& \text{Treatment B} \\ \hline \text{Small stones} & \text{Group 1}& \text{Group 2} \\ & \text{93%}(\frac{81}{87}) & \text{87%}{(\frac{234}{270})}\\ \hline \text{Large stones} & \text{Group 3}& \text{Group 4}\\ & \text{73%}(\frac{192}{263)} & \text{69%}{(\frac{55}{80})}\\ \hline \text{Both} & 78\%(\frac{273}{350}) &\text{83%}(\frac{289}{350})\\ \hline \end{array}$$
Suppose a patient is randomly selected. Let $S$ := Patient has small stones, $C$ := The treatment was successful, and $A$ := The patient was given Treatment A. Using the following result (*) and conditioning on $S$, verify the values for $\Pr \left( {C \mid A} \right)$ and $\Pr \left( {C \mid {A^c}} \right)$.
(*) Let $\left\{ {{B_1},{B_2}, \ldots } \right\}$ be events that are pairwise mutually exclusive and exhaustive, and $A$, $C$ be arbitrary events. Then: $$\Pr \left( {A \mid C} \right) = \sum\limits_{n = 1}^\infty {\Pr \left( {A \mid {B_n} \cap C} \right)\Pr \left( {{B_n} \mid C} \right)} $$
So my confusion is:
In the case of finding $\Pr \left( {C \mid A} \right)$, is the following the right use of the result (*) with ''conditioning on $S$''? $$\begin{gathered} \Pr \left( {C \mid A} \right) & = \Pr \left( {C \mid \left( {S \cap A} \right)} \right) \cdot \Pr \left( {S \mid A} \right) + \Pr \left( {C \mid \left( {{S^c} \cap A} \right)} \right) \cdot \Pr \left( {{S^c} \mid A} \right) \hfill \\ & = \frac{{\Pr \left( {C \cap S \cap A} \right)}}{{\Pr \left( {S \cap A} \right)}} \cdot \frac{{\Pr \left( {S \cap A} \right)}}{{\Pr \left( A \right)}} + \frac{{\Pr \left( {C \cap {S^c} \cap A} \right)}}{{\Pr \left( {{S^c} \cap A} \right)}} \cdot \frac{{\Pr \left( {{S^c} \cap A} \right)}}{{\Pr \left( A \right)}} \hfill \\ \end{gathered} $$
Yes, that's correct, as you can see by simplifying the last expression.
$$\frac{{\Pr \left( {C \cap S \cap A} \right)}}{{\Pr \left( {S \cap A} \right)}} \cdot \frac{{\Pr \left( {S \cap A} \right)}}{{\Pr \left( A \right)}} + \frac{{\Pr \left( {C \cap {S^c} \cap A} \right)}}{{\Pr \left( {{S^c} \cap A} \right)}} \cdot \frac{{\Pr \left( {{S^c} \cap A} \right)}}{{\Pr \left( A \right)}} \\=\frac{\Pr(C\cap S\cap A)}{\Pr(A)}+\frac{\Pr(C\cap S^c\cap A)}{\Pr(A)} \\=\frac{\Pr((C\cap S\cap A)\cup(C\cap S^c\cap A))}{\Pr(A)} \\=\frac{\Pr(C\cap A)}{\Pr(A)}=\Pr(C|A) $$