These probabilities are not required to sum to 1, because they are not incompatible and exhaustive possible outcomes of an experiment. More obvious example to illustrate:
P(6-sided die coming up as 6 | today is Monday) = 1⁄6 P(6-sided die coming up as 6 | today is not Monday) = 1⁄6 1⁄6 + 1⁄6 != 1
I think your example is not suitable for situation above—there I can see only two possible outcomes: X happen or X not happen. We don’t know anything more about X. And P(X|A) + P(X|~A) = 1, isn’t so?
Yes, either X happens or X doesn’t happen. P(X) + P(~X) = 1, so therefore P(X | A) + P(~X | A) = 1. Both formulations are stating the probability of X. But one is adjusting for the probability of X given A; so either X given A happens or X given A doesn’t happen (which is P(~X | A) not P(X | ~A)).
These probabilities are not required to sum to 1, because they are not incompatible and exhaustive possible outcomes of an experiment. More obvious example to illustrate:
P(6-sided die coming up as 6 | today is Monday) = 1⁄6
P(6-sided die coming up as 6 | today is not Monday) = 1⁄6
1⁄6 + 1⁄6 != 1
I think your example is not suitable for situation above—there I can see only two possible outcomes: X happen or X not happen. We don’t know anything more about X. And P(X|A) + P(X|~A) = 1, isn’t so?
No. You may have confused it with P(X|A) + P(~X|A) = 1 (note the tilda). In my case, either 6-sided die comes up as 6, or it doesn’t.
Yes, either X happens or X doesn’t happen. P(X) + P(~X) = 1, so therefore P(X | A) + P(~X | A) = 1. Both formulations are stating the probability of X. But one is adjusting for the probability of X given A; so either X given A happens or X given A doesn’t happen (which is P(~X | A) not P(X | ~A)).