# Is this an instance of the base-rate fallacy?

The following line of probability reasoning is supposedly fallacious, and is an instance of the base-rate fallacy. The argument is that $(1)-(3)$ don't give us enough reason to conclude that $(C)$.

But it seems to me that this is not the case. I can only see how (C) fails to follow from (2)-(3). That is, admitting (1) forces (C) to be true given (2)-(3). So am I correct, or is the following indeed fallacious reasoning?

$$\tag{1} Pr(Sx \mid x \in \mathcal{H}) \gg 0$$

$$\tag{2} Pr(Sx \mid Tx \wedge x \in \mathcal{H}) \gg 0$$

$$\tag{3} Pr(Sx \mid \neg Tx \wedge x \in \mathcal{H}) \ll 1$$

$$\tag{C} Pr(Tx \mid Sx \wedge x \in \mathcal{H}) \gg 0$$

I think you are correct, unless I have misunderstood something in your notation. I made an elementary probability calculation writing $A$ for $Sx$, $B$ for $x \in \mathcal{H}$ and $C$ for $Tx$. Then assuming all probabilities are nonzero:
You assume that $P(A | \neg C \cap B)$ is small and $P(A|B)$ is large, and so $P(C | A \cap B)$ must be close to $1$.
• Yes, it seems that (2) follows from (1) and (3) as follows: $P(A|B \cap C) = P(A\cap B \cap C)/P(B \cap C) = (P(A\cap B) - P(A \cap \lnot C \cap B))/P(B \cap C) \approxeq P(A\cap B)/P(B \cap C) = P(A|B) /P(C|B) \ge P(A|B) \approxeq 1$ by (1). – Flounderer May 6 '15 at 22:13