What is $1 - P(H|X)$ where $P(H|X)$ is conditional probablity?

Question

I understand that if $P(X)$ is the probability of $X$ happening, then $1 - P(X)$ is the probability of $X$ not happening.

However I am unable to understand how does this play out when we deal with conditional probabilities.

So $P(H|X)$ is probability of $H$ happening given $X$ has already happened, what's $1 - P(H|X)$?

Answer 1

Let $X = \{1,2,3,4,5\}$, and $H$ choose 1,2, or 3. All choices are made with equal chance. Then $P(H|X)$ is that chance that you chose 1,2, or 3, and $1-P(H|X)$ is the chance that you chose 4 or 5.

You could imagine that $X$ is a subset of a larger set like $Y = \{1,\dotsc, 100\}$.

Answer 2

Use the definition of conditional probability:

$$P(H|X)=\frac{P(H\cap X)}{P(X)}.$$

Then

$$1-P(H|X)=1-\frac{P(H\cap X)}{P(X)}=\frac{P(X)-P(H\cap X)}{P(X)}.$$