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According to this video, one can freely decide to conceptualize probabilities in terms of either event language or proposition language. It states, "the mathematical rules are applied the same either way; it's just a matter of how you interpret the language."

I have been running into the claim that when $X$ is an event, $X\ |\ A$ is not an event. Rather, the "$|\ A$" in the expression $\Pr(X\ |\ A)$ is taken to be part of the $\Pr()$ function (that is, $\Pr(X\ |\ A)$ should be interpreted as $\color{blue}{\Pr(}\color{red}X\ \color{blue}{|\ A)}$ instead of $\color{blue}{\Pr(}\color{red}{X\ |\ A}\color{blue})$).

However, this doesn't sit quite right with me, because when $X$ is a proposition, $X\ |\ A$ is clearly a proposition. For example, if $X$ is the proposition, "I will use an umbrella," and $A$ is the proposition, "it will rain," then $X\ |\ A$ becomes "I will use an umbrella given that it will rain." This is the conditional statement, "if it rains, then I will use an umbrella," which can be assigned a truth value using propositional logic, and is thus a proposition. This pretty clearly indicates that in this case, $X\ |\ A$ does have intrinsic meaning apart from the $\Pr()$ function, and returns the same object type as $X$ (in this case, a proposition). This is useful for computations, as it allows, for example, substitution (i.e., $Y = X\ |\ A$).

Does this truly constitute mathematics that can only be done in proposition language, and not in event language, thus violating the claim that the distinction between the two is merely interpretive? Is it possible to port this mathematical utility back over to event language? If not, assuming every problem can be formulated in either language, would it make sense to migrate to talking in proposition language more often and slowly phase out event language?

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Events are sets of outcomes in an outcome space. Probabilities are measures of these sets. Conditional probabilities are relative measures. By definition:$$\mathsf P(U\mid R)=\dfrac{\mathsf P(R\cap U)}{\mathsf P(R)}$$

This conditional probability measures the event "It rains and I use an umbrella" against the event of "It rains".

The event of "If it rains, then I will use an umbrella," is the event of "It does not rain or I use an umbrella"

So, no. Clearly $\mathsf P(R^\complement\cup U)$ and $\mathsf P(U\mid R)$ are not measuring the same things.

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