# contrapositive of probability

If P(A|B) = 95%, then is P(B'|A') also 95%?

The subject is hypothesis testing. If the null hypothesis is true and there is a 95% probability that the data should pass the test, then does failing the test imply the null hypothesis is wrong with 95% chance?

• Thanks to all the wonderful answers. Each one added something important to the question! The question is sure fraught with ambiguity and loose patterns, sometimes combining incompatible ideas that are well-defined in common statistics language. The nugget of concreteness within was exposed through algebra and some good reasoning about upper and lower bounds. Mar 20, 2013 at 3:46

No, $P(A\mid B) = 0.95$ does not tell you very much about $P(B^c\mid A^c)$.

For example, suppose that $P(B) = 0.5$ and $P(A\cap B) = 0.475$ so that $$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{0.475}{0.5} = 0.95.$$ We also have that $P(A^c\cap B) = P(B) - P(A\cap B) = 0.5-0.475 = 0.025$. Now consider two possibilities for $P(A^c\cap B^c)$ and $P(A\cap B^c)$.

• Suppose $P(A^c\cap B^c) = 0$ and $P(A\cap B^c) = 0.5$.
This means that $P(A) = P(A\cap B) + P(A\cap B^c) = 0.475 + 0.5 = 0.975$ and so $P(A^c) = 1-P(A) = 0.025$. Hence, $$P(B^c\mid A^c) = \frac{P(A^c \cap B^c)}{P(A^c)} = 0.$$

• Suppose $P(A^c\cap B^c) = 0.5$ and $P(A\cap B^c) = 0$.
This means that $P(A^c) = P(A^c\cap B) + P(A^c\cap B^c) = 0.025+0.5 = 0.525$. Hence, $$P(B^c\mid A^c) = \frac{P(A^c \cap B^c)}{P(A^c)} = \frac{0.5}{0.525} = 0.95238\ldots.$$

For intermediate choices of $P(A^c\cap B^c)$ and $P(A\cap B^c)$, we can come up with other values, including your desired $0.95$, for $P(B^c\mid A^c)$.

• If P(A')=P(B), then P(B'|A') = P(A|B). Wait.. I'm not sure just yet. Mar 19, 2013 at 20:08
• Ok, I'm reasonably confident in this: If P(A' n B')=P(A n B), then P(B'|A') = P(A|B). Mar 19, 2013 at 20:14
• @TroyofHelen Nice! If $P(AB) = p = P(A^cB^c)$ and $P(AB^c)=q, P(A^cB)=1-q-2p$, then I get $P(A)=p+q$, $P(A^c) = 1-p-q$ and $P(B)=1-p-q$ giving $$P(B^c\mid A^c) = \frac{P(A^cB^c)}{P(A^c)} = \frac{p}{1-p-q} = P(A\mid B) = \frac{P(AB)}{P(B)} = \frac{p}{1-p-q}.$$ Also, interestingly enough, $P(A^c) = P(B)$. Mar 19, 2013 at 22:24
• huh, I thought the contrapositive was important for some kind of symmetry, but with probabilities, it really isn't necesary. So why not find out what is needed for P(A|B) = P(B|A)? --- Okay, P(B)=P(A) is a one-line step. What does this mean? Maybe the two events are identical. ---- Now here's some loose reasoning that I find intuitive -> ---- Is picking a hypothesis the same kind of thing as recording a test statistic? If so, then the confidence we have in the hypothesis given the data is the same as the chance of the data given the hypothesis. That seems right at first glance. Mar 20, 2013 at 3:35

To approach it conceptually, I'd say that the contrapositive is only implied in the case of absolute conditionals. A->B means "A always implies B (i.e., P(B|A) = 1). It does not mean "this conditional is considered correct for any case in which A and B are true." If P(B|A) = .95 then there are cases that contradict the statement A->B and therefore the contrapositive of that rule is not implied in any fashion.

• Generally, contrapositive is indeed only absolute! There could be a condition that is needed for this contrapositive of probability to work. I don't know if this is anything formalized or if there are similar ideas. Mar 19, 2013 at 20:16