Question #51953

A number $x$ is chosen at random from the set $S = \{1, 2, \dots, 361\}$. Let $A$, $B$, and $C$ be the events that $x$ is divisible by 2, 3, and 19, respectively. A diagnostic test is performed on $x$ according to the following rules: \begin{itemize} \item If $x \in C$, the test yields a positive result with probability $\frac{10}{19}$. \item If $x \notin C$ and $x \in A^c \cap B^c$, the test yields a positive result with probability $\frac{5}{19}$. \item If $x \notin C$ and $x \notin A^c \cap B^c$, the test yields a positive result with probability $\frac{7}{19}$. \end{itemize} Given that the test yields a positive result, what is the probability that $x$ is NOT divisible by any of 2, 3, or 19?