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?
 A: 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 $P(B^c\mid A^c)$.
A: 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. 
A: A couple of points in addition to the good answers you see already.
You use the phrase "null hypothesis" which is generally used in frequentist statistics.  In frequentist statistics the null hypothesis is a fixed fact not subject to probability so saying something like the "null hypothesis is wrong with 95% chance" is meaningless.
Bayesians define probability in terms of our knowledge about something and can therefore talk about the probability of a null hypothesis being true, but most Bayesians don't like to use the phrase "null hypothesis".  But if we mix the 2 and talk about a Bayesian probability of the null hypothesis being true, then we also need a prior distribution and the posterior probability will depend on that prior along with the data.
Consider the null hypothesis that a coin is 2-headed (prob of heads is 1) and the observed data of 1 flip of the coin which came up heads.  In frequentist statistics this would result in a p-value of 1 (or 100%) which means that the observed data is consistent with the null hypothesis.  It does not mean that there is a 100% chance that the coin is 2 headed.  In fact the data we have is consistent with a null of a fair coin (p=0.5) and other possibilities as well.  If we only do this once then the coin is either 2-headed or it is not, there is no probability involved with regaurd to the coin itself.
