# 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.

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

• Ah, I see! Thank you :) – ConfusedJimbo Mar 17 '16 at 9:31

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

• Ah, I see! Thank you :) – ConfusedJimbo Mar 17 '16 at 9:31