Is it true that $P(A|B)+P(A|\overline B) = 1$? tl;dr
$P(A|B) + P(A|\overline B)=1$. My question is, is this true?
More detail
The book I'm reading (Statistics for Business and Economics, Paul Newbold et al) has this example (paraphrased a little).

10% athletes have used performance enhancing drugs. A test is available that correctly identifies an athlete's drug usage 90% of the time. If an athlete is a drug user, the probability is 0.9 that the athlete is correctly identified by the test as a drug user. Similarly is not a drug user, the prob is 0.9 that the athlete is correctly identified as not using drugs.

So this is fine so far. From this I'd get the following:
Let D = athlete is a drug user, + = test calls positive, - = test calls negative
$ \begin{align}
   P(D) &= 0.10  \implies P(\overline D)\\
   P(+|D) &= 0.90 \\
   P(-|\overline D) &= 0.90
  \end{align}
$
Where the example looses me is that it says that the following can also be defined from the above information...
$ \begin{align}
   P(+|\overline D) &= 0.10 \\
   P(-|D) &= 0.10
  \end{align}
$
And this is where I'm struggling... it seems that they can only be getting this by assuming $P(+|D)+P(+|\overline D)=1$, or more generally $P(A|B) + P(A|\overline B)=1$. My question is, is this true?
Here's what I've tried so far to see if it is true...
$
\begin{align}
P(A|B)+P(A|\overline B) &= \frac{P(A \cap B)}{P(B)} + \frac{P(A \cap \overline B)}{P(\overline B)}\\
&= \frac{P(A \cap B)P(\overline B) + P(A \cap \overline B)P(B)}{P(B)P(\overline B)} \\
&= \frac{P(A \cap B)(1-P(B)) + P(A \cap \overline B)P(B)}{P(B)P(\overline B)}\\
&= \frac{P(A \cap B) - P(A \cap B)P(B) + P(A \cap \overline B)P(B)}{P(B)P(\overline B)}\\
&= \frac{P(A \cap B)}{P(B)(1 - P(B))} + \frac{P(A \cap \overline B) - P(A \cap B)}{P(\overline B)} \\
&= \frac{P(A \cap B)}{P(B)(1 - P(B))} + \frac{P(A \cap \overline B) - (1 - P(A \cap \overline B))}{P(\overline B)} \\
&= \frac{P(A \cap B)}{P(B)(1 - P(B))} + \frac{2\cdot P(A \cap \overline B) - 1}{P(\overline B)} \\
&= \frac{P(A \cap B)}{P^2(B)} + \frac{2\cdot P(A \cap \overline B) - 1}{P(\overline B)} \\
&= ?????
\end{align} 
$
And then stuck... can't see how this is true. And looking at a Venn diagram representation it didn't enlighten me... any thoughts on how this question's answer in the book is making the assertions it is? Many thanks.
 A: No, that's not what the author is using, it is wrong in general, as you saw. What he is doing, is the following 
$$ \def\P{\mathbf P} \P[\bar A \mid B] + \P[A \mid B] = 1 $$
or in your case
$$ \P[+ \mid \bar D] = 1 - \P[-\mid \bar D] = 1 - \frac{9}{10} = \frac 1{10} $$
A: Why would this be true?
If I roll a die, then the probability that I am a hamster given that I roll an even number is $0$, and the probability that I am a hamster given that I roll an odd number is also $0$.
What you do have is $$P(A|B) + P(\bar A|B) = 1$$
What you do not have is $$P(A|B)+P(A|\bar B) = 1$$
A: First, I want to introduce the proof of Law of total probability:
\begin{align}
P(A \mid B) P(B) + P(A \mid \bar B) P(\bar B)
&= \frac{P(A \cap B)}{P(B)} P(B) + \frac{P(A \cap \bar B)}{P(\bar B)} P(\bar B) \\
&= P(A \cap B) + P(A \cap \bar B) \\
&= P[(A \cap B) \cup (A \cap \bar B)] + P[(A \cap B) \cap (A \cap \bar B)] \\
&= P(A) + P(\varnothing) \\
&= P(A) + 0 \\
&= P(A)
\end{align}

Let $P(B) = p$, $P(\bar B) = 1 - p$, $P(A) = q$, $P(A \mid B) = x$, $P(A \mid \bar B) = y$, we can form the following linear equation:
\begin{align}
\left\{
\begin{array}{l}
p x + (1-p) y = q \\
x + y = 1
\end{array}
\right.
\end{align}
We can solve it:
\begin{align}
x &= \frac{\begin{vmatrix}
q & 1 - p \\
1 & 1
\end{vmatrix}}
{\begin{vmatrix}
p & 1 - p \\
1 & 1
\end{vmatrix}}
= \frac{p + q - 1}{2p - 1}
\\
y &= \frac{\begin{vmatrix}
p & q \\
1 & 1
\end{vmatrix}}
{\begin{vmatrix}
p & 1 - p \\
1 & 1
\end{vmatrix}}
= \frac{p - q}{2p - 1}
\end{align}
In conclusion, $P(A \mid B) + P(A \mid \bar B) = 1$ does NOT hold unless the equation above holds.
