Clinical trials with $0$ sensitivity A probability problem which seems interesting (and I'm not sure about the answer):
Two companies A and B are developing a test for a disease (affects $1\%$ of population). Company A's test has specificity sensitivity $0.95$.
Company B does a trick, every patient is diagnosed as not having the disease. 
My questions:
a). For the test developed by B, sensitivity is equal to 0 (specificity $=1$)? 
b). How do we define overall success mathematically (a random person is correctly diagnosed)?
c). Is the overall success higher for test B vs test A?   
For a) I want to be sure that it is so. For b) I am inclined to say that overall success is given by the $P(T,D)+P(T^c,D^c)$, where $T=\text{test is positive}$ and $D=\text{has the disease}$ and I'm not sure (I think this is case for Simpson's paradox if we take into account only the success rate).
Sensitivity is defined as: 
$P(T|D)=0.95$.
Specificity is defined as:
$P(T^C|D^C)=0.95$
 A: a) I don't know what specificity and sensitivity mean exactly. Most online sources give fractions. Some give that
$$\text{sensitivity} = P(+|D)$$
where let $+$ mean a positive result, and $D$ is the event that a patient has the disease. If this is the case, then if I tell you that a particular patient has the disease, then test B will always say negative, hence
$$P(+|D) = 0$$
Similarly, I found
$$\text{specificity} = P(-|\bar D)$$
where $-$ is the event that the test says negative, and $\bar D$ that the patient does not have it. If I tell you that a particular person does not have the disease, then test B will always say negative, hence
$$P(-|D) = 1$$
So I agree.
b) I'm not sure either, but I would venture to say that we must compare the probability of accuracy for either test. For example, if I let $\mathscr A$ be the event that test A was accurate,  $T$ the test was positive, $\bar T$ the test was negative, then the probability that test A was accurate is
$$P(\mathscr A) = P(TD)+P(\bar T\bar D) = P(T|D)P(D)+P(\bar T|\bar D)P(\bar D) = .95(.01)+.95(.99)=.95 $$
where the second equality is due to the law of total probability, and in the second equality I invoked the product rule. Also, I assume you meant that sensitivity = specificity = .95
Similarly, we have
$$P(\mathscr B) = P(TD)+P(\bar T\bar D) = P(T|D)P(D)+P(\bar T|\bar D)P(\bar D) = 0(.01)+1(.99) = .99$$
c) What do you think? Why is this not surprising? What happens if instead the the disease affects $10\%$ of the population?
(Rhetorical.) 
