Given $P(C)$ and $P(A\mid C)$, what is $P(C\mid A)$? I am wondering if there's a way to find the solution if we know:
$P(C) = 0.01$
$P(A\mid C) = 0.7$
what is $P(C\mid A)$?
I think we need to know $P(A)$ to answer this question right? There is no other way around it?
Thank you!

\begin{align}
\Pr(A) & = \Pr(A\mid C)\Pr(C) + \Pr(A\mid \text{not }C)\Pr(\text{not }C) \\[10pt]
& = \Pr(A\mid C)\Pr(C) + \Pr(A\mid \text{not }C)\Big( 1 - \Pr(C) \Big) \\[10pt]
& = (0.7)(0.01) + \Pr(A\mid\text{not }C)( 1 - 0.01) \\[10pt]
& = 0.007 + 0.99\Pr(A \mid \text{not } C) \\[10pt]
& \in [0.007,\, 0.997]
\end{align}
Since the probability of $C$ is so small, this range at the end is very wide.
So
$$
1 \ge \frac{0.007}{\Pr(A)} \ge \frac{0.007}{0.997} \approx 0.007021063\ldots
$$
