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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.

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

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  • $\begingroup$ Ah, I see! Thank you :) $\endgroup$ – ConfusedJimbo Mar 17 '16 at 9:31
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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 $$P(A|B)+P(A|\bar B) = 1$$

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

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