Does this reasoning using a combination of contraposition and probabilty work? We can write "predicate $P$ implies predicate $Q$", (AKA "100% of the time, when $P$ occurs, $Q$ occurs",) like so:
$$P \implies Q$$
Analogously, we can write "90% of the time, when $P$ occurs, $Q$ occurs" as $$P \overset{\mbox{90%}}{\implies} Q$$
If we have some set of undesirable (bad), possible events $B$, and we have noticed empirically that around 90% of the time, a predicate $T$ which we can test for, is false, i.e. given a $b \in B$, we can say $$b \overset{\mbox{90%}}{\implies} \neg T$$
With non-probabilistic implications, e.g. $P \implies Q$ we can use contraposition to infer $\neg Q \implies \neg P$
Given the above definitions, and the statement $b \overset{\mbox{90%}}{\implies} \neg T$, it seems like we can infer the following: $$T \overset{\mbox{90%}}{\implies} \neg b$$
which I interpret as "If we ensure predicate $T$ is true, then we have prevented 90% of the events in $B$".
Is this inference valid? If not, why not?
 A: No, it isn't valid. If $b$ implies not-$T$ always, then there is no overlap between $b$ and $T$, so we can indeed conclude that $T$ implies not-$b$. If $b$ implies not-$T$ $90$% of the time, it means there is an overlap between $b$ and $T$ which covers $10$% of $b$, but without more information we have no idea what percentage the overlap is of $T$.
A: Your question can be rephrased in conditional probability terms:

Suppose I know $P(\neg T | b) = 0.9$. Must it be true that $P(\neg b | T) = 0.9$?

The answer is "no"; even if we are given $P(\neg T | b) = 0.9$, $P(\neg b | T)$ can take any value between $0$ and $1$, depending on the individual likelihoods of $b$ and $T$.
Let's take a specific example. Suppose I take a blood test for a disease. This blood test gives me either a "not bad" ($\neg b$) or "bad" ($b$) result. "Not bad" here means "Test says disease free"; "bad" means "Test says diseased".
But this blood test also has a very high rate of false positives - a 90% rate, in fact. So if I get a "bad" test result, there is a 90% chance that it is not true ($\neg T$) that I have the disease, and a 10% chance that it is true ($T$) that I have the disease. This is equivalent to $P(b | \neg T) = 0.9$.
Stated in this concrete instance, it seems unlikely that the probability $P(T | \neg b)$ that my test result is "not bad" given that it is "true" that I have the disease, must be 90%. And indeed, by picking different values for $P(b)$ and $P(T)$ overall, I can get a range of numbers for this conditional probability.
For instance, suppose $P(b) = 0.2$ (a 20% likelihood of getting a "bad" result) and $P(T) = 0.04$ (a 4% likelihood of truly having the disease). Then if $P(\neg T|b) = 0.9$, it must be that $P(b \cap \neg T) = P(\neg T | b) * P(b) = 0.9 * 0.2 = 0.18$ and $P(b \cap T) = P(b) - P(b \cap \neg T) = 0.20 - 0.18 = 0.02$, from which we quickly find $P(\neg b \cap T) = 0.02$ and $P(\neg b \cap \neg T) = 0.78$:
\begin{align*}
  && T \text{ (infected)} && \neg T \text{ (healthy)} && \text{Total} \\
\neg b \text{ (not bad)} && 0.02 && 0.78 && 0.80 \\
b \text{ (bad)} && 0.02 && 0.18 && 0.20 \\
\text{Total} && 0.04 && 0.96 && 1.00 \
\end{align*}
So in this instance, $$P(\neg b | T) = \frac{P(\neg b \cap T)}{P(T)} = \frac{0.02}{0.04} = 0.5,$$ i.e. I have a 50% chance of getting a negative result if I am truly infected, instead of a 90% chance.
Now let's change those numbers to see that $P(\neg b | T)$ also changes. If I take $P(b) = 0.30$, $P(T) = 0.05$, then I find $P(b \cap \neg T) = 0.9 * 0.30 = 0.27,$ $P(b \cap T) = 0.03$, $P(\neg b \cap T) = P(T) - P(b \cap T) = 0.05 - 0.03 = 0.02:$
\begin{align*}
  && T \text{ (infected)} && \neg T \text{ (healthy)} && \text{Total} \\
\neg b \text{ (not bad)} && 0.02 && 0.68 && 0.70 \\
b \text{ (bad)} && 0.03 && 0.27 && 0.30 \\
\text{Total} && 0.05 && 0.95 && 1.00, \
\end{align*}
and now we have that $$P(\neg b | T) = \frac{P(\neg b \cap T)}{P(T)} = \frac{0.02}{0.05} = 0.4,$$ i.e. I have a 40% chance of getting a negative result if I am truly infected, instead of a 90% chance.
We can express $P(\neg b | T)$ as a function of $x = P(b)$ and $y = P(T)$ as follows. Since events can't have negative probability, we know that $P(T \cap b) \leq P(T)$ and $P(T \cap b) = 0.1x$, meaning that $y \geq 0.1x$. Any numbers $0 \leq x, y \leq 1$ satisfying the constraint $y \geq 0.1x$ can make sense in the problem. Then the conditional probability $P(\neg b | T)$ is equal to $$P(\neg b | T) = \frac{P(\neg b \cap T)}{P(T)} = \frac{y - 0.1x}{y},$$ which can clearly take on the entire range of values $(0, 1)$.
