# How does P(T|D) + P(T|~D) = 1?

I was reading the book I came across the conditional probability. The statement is:

Imagine a certain disease that affects 1 in every 10,000 people. And imagine that there is a test for this disease that gives the correct result (“diseased” if you have the disease, “nondiseased” if you don’t) 99% of the time.

T = Test result is positive. D = Person is diseased.

So, P(T|D) = 0.99 and P(D) = 0.0001. The author calculated P(D|T) by Bayes theorem.

P(D|T) = P(T|D) P(D) / [P(T|D)P(D) + P(T|~D)P(~D)]

For that, he calculated different probabilities.i.e.

• P(~D)= 1- P(D) = 0.9999
• P(T|~D) = 1 - P(T|D) = 0.01

I could understand intuitively that P(~T|D) = 1 - P(T|D). But I am unable to understand how P(T|~D) = 1 - P(T|D) is valid? Can somebody explain mathematically?

P.S. The snapshot of the book snippet is: • ... It isn't true in general; in this case the "gives the correct result (...) 99% of the time" sentence makes two distinct things the same, which is why the result holds in this particular case Jun 4, 2017 at 11:04

In general, P(T|~D) ≠ 1 - P(T|D). To see why, suppose (1) the test results were completely independent of the disease and (2) the test is positive 25% of the time.

(1) implies that $$P(T|\lnot D) = P(T|D) = P(T).$$ If the test results are completely independent of the disease, then it doesn't matter what the disease state is. Then the equation you're asking about becomes $$P(T|\lnot D) =^? 1 - P(T|D)$$ $$P(T) =^? 1 - P(T)$$ The second version is true if, and only if, P(T) = .5. In particular, filling in the value from (2), $$.25 ≠ 1 - .25 = .75.$$

It looks like the book section you were reading is here. I don't see where the author is assuming that P(T|~D) = 1 - P(T|D).

Maybe you're getting confused by the way the author moves between two versions of the total probability. One version is $$P(F) = P(F,E) + P(F,\lnot E).$$ The second is $$P(F) = P(F|E)P(E) + P(F|\lnot E)P(\lnot E).$$

The first version expresses the probability of F in terms of two conjunctions: F-and-E, F-and-not-E. The second version expresses the probability of F in terms of two conditional probabilities: F-given-E and F-given-not-E. The definition of conditional probability makes them equivalent. But it's important to note that the conditional version includes the marginal (unconditional) probabilities of E and ~E.

Another potential point of confusion is that the author does most of the calculation in non-percentage numbers (.99 rather than 99%), then reports the final value as a percentage (.98% rather than .0098).

• I inferred P(T|~D)= 1 - P(T|D) because the author has calculated P(T|~D)=0.01 and P(T|D)=0.99. So, I became confused that how he calculated P(T|~D)=0.01. Jun 4, 2017 at 11:57
• The author assumes that the test "gives the correct result (“diseased” if you have the disease, “nondiseased” if you don’t) 99% of the time." That is, P(T|D) = .99 and P(~T|~D) = .99. So the equation they use is P(T|~D) = 1 - P(~T|~D). Jun 4, 2017 at 12:19