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I had a question regarding the properties of probability.

If we were to say that we have "any two events $A$ and $B$," is it okay to assume that they both belong to the same sample space $S$ (i.e. $A \subseteq S$ and $B \subseteq S$)?

I'll give a specific example exercise that prompted me to ask this question:

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"Show that for any events $A$ and $B$, $\ $ $P(A) + P(B) - 1 \le P(A \cap B)$."

What I did is move the 1 over, so now we have

$$P(A) + P(B) \le 1 + P(A \cap B)$$

which is true if we assume that $P(A) + P(B) \le P(S) = 1$.

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I hope my question makes sense. Thank you for the feedback!

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    $\begingroup$ The question certainly intends for you to assume that $A$ and $B$ are subsets of the same sample space, but perhaps this (or something to this effect) should be stated explicitly in the question. I'd be interested in hearing suggestions for a more precise and yet non-cumbersome way of phrasing the example question. $\endgroup$ – littleO Sep 14 '18 at 7:40
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    $\begingroup$ I may misunderstood you, but do you claim that assuming $A, B\subseteq S$ is equivalent with $P(A) +P(B) \leq P(S) $? $\endgroup$ – Shashi Sep 14 '18 at 7:46
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    $\begingroup$ For your example exercise you only need to apply inclusion-exclusion. Practically I do not see people will explicitly stating this as it is obvious(?). If there is a probability measure $P$ such that $P(A), P(B)$ are well-defined, there is also a corresponding sample space $\Omega$ and sigma-algebra $\mathcal{F}$ to form a probability space, and by definition $A, B \subseteq \Omega$ and $A, B \in \mathcal{F}$. $\endgroup$ – BGM Sep 14 '18 at 7:52
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    $\begingroup$ $P(A)+P(B)\le1+P(A\cap B)$ is true, but $P(A)+P(B)\le1$ is not necessarily true. For example, if $A=B=S,$ then $P(A)+P(B)=2.$ $\endgroup$ – bof Sep 14 '18 at 8:21
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    $\begingroup$ @Sean You are very confused. Probabilities can't exceed one. The sum of two probabilities can be as big as 2. Let's look at a concrete example. The random experiment is rolling a fair $6$-sided die. The sample space $S$ consists of the $6$ outcomes $1,2,3,4,5,6.$ Let $A$ be the event $X\le3$ and $B$ the event $X\ne2.$ Then $P(A)=\frac12$, $P(B)=\frac56$,and $P(A)+P(B)=\frac12+\frac56=\frac43\gt1.$ On the other hand, $P(A\cup B)=1.$ $\endgroup$ – bof Sep 14 '18 at 10:55
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In this case it's implicit in the notation.

Recall the definition. A probability space is a triple: $ \left(S, \mathcal F, \mathbb P\right)$ where $\mathcal F$ is a set of "events" (Subsets of $S$) and $\mathbb P$ is a function from $\mathcal F$ to $\left[ 0,1 \right]$.

So the expression $$\mathbb P(A) + \mathbb P(B) - 1 \leq \mathbb P(A\cap B)$$ only makes sense if $A, B \in \mathcal F$ The function $\mathbb P$ carries its domain around with it, meaning that it can only ever refer to one sample space.

In addition the term event is ambiguous if there are two sample spaces involved. So the use of the term event usually means that there is only one sample space under consideration. Which means that "Any two events $A$ and $B$" can be taken to imply that $A$ and $B$ are events in the same sample space.

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