Is it true that $P(A|B)+P(A|\overline B) = 1$?

Question

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.

Answer 1

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 $$ \P[+ \mid \bar D] = 1 - \P[-\mid \bar D] = 1 - \frac{9}{10} = \frac 1{10} $$

Answer 2

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 $$P(A|B)+P(A|\bar B) = 1$$

Answer 3

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.