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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?

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

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

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

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

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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:

\begin{align*}P(C | A \cap B) = 1 - P(\neg C | A \cap B) &= 1-P(\neg C \cap A \cap B)/P(A \cap B)\\ &= 1-\frac{P(A | \neg C \cap B) P(\neg C \cap B)}{P(A \cap B)}\\ &= 1-\frac{P(A | \neg C \cap B) (P(B) - P(C \cap B))}{P(A \cap B)}\\ &= 1-\frac{P(A | \neg C \cap B) (1 - P(C | B))}{P(A | B)}\\ &\ge 1 - \frac{P(A | \neg C \cap B)}{P(A|B)} \end{align*}

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 $1$.

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  • $\begingroup$ This is very elucidating, and it looks like (2) isn't being used at all. Instead, one only needs (1) and (3) to show that (C) using my notation above. Is this correct? $\endgroup$ – PP121 May 6 '15 at 16:50
  • $\begingroup$ 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). $\endgroup$ – Flounderer May 6 '15 at 22:13

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