Question about a sample space

Question

I am reading a book on probability theory, and i stumbled a little on this question:

A sample space contains two events $E$ and $F$, and $P(E) = 0.70$, $P(F) = 0.25$, $P(E \cap F) = 0.15$

Determine $P(\bar{E})$, $P(E \cup F)$, ...

So i have 2 questions:

1. Why does their probability not add up to 1? Are they just assuming that other events likely also exist in this sample space, that they just didn't mention?
2. When they define $P(E \cap F) = 0.15$, would that probability $0.15$ be counted towards the entire sample space, or is it already counted in the definition of $P(E)$ and $P(F)$, because i feel like we already counted those probabilities.

Answer 1

Think of the probability of event $E$ as broken into two parts: the part that overlaps with event $F$ and the part that is distinct from event $F$. We capture this be noting that the part that overlaps is just the intersection of the two events, $E \cap F$, whereas the second part that is distinct from $F$ is the same as the intersection of event $E$ and everything NOT in $F$, i.e., the complement of $F$, or $\overline F$. Thus, event $E$ is really $$E = (E \cap F) \cup (E \cap \overline F).$$ Now, the right-hand side is disjoint (no overlap), so we can write $$P(E) = P(E \cap F) + P(E \cap \overline F).$$ With a bit of rearranging, we get $$P(E \cap \overline F) = P(E) - P(E \cap F)$$ and since we have the values for the two probabilities on the right, we can find the probability of the part of event $E$ that does not overlap with event $F$.

Of course, there is nothing special about the roles of $E$ and $F$, so we could switch them to get the corresponding relationship for event $F$: $$F = (F \cap E) \cup (F \cap \overline E).$$ Which gives $$P(F) = P(F \cap E) + P(F \cap \overline E)$$ and $$P(F \cap \overline E) = P(F) - P(E \cap F)$$

Now, the union of the two events can be broken into three parts, the intersection of the two events, and the parts that are distinct to each event (as above) $$E \cup F = (E \cap \overline F) \cup (E \cap F) \cup (\overline E \cap F).$$ The nice part of the right-hand side is that all three of the events in the parentheses are disjoint...there is no overlap. Thus $$P(E \cup F) = P(E \cap \overline F) + P(E \cap F) + P(\overline E \cap F).$$

Now, if we rewrite this as $$P(E \cup F) = P(E \cap \overline F) + P(E \cap F) + P(\overline E \cap F) + P(E \cap F) - P(E \cap F)$$ we get $$P(E \cup F) = P(E) + P(F) - P(E \cap F)$$ Having put these together, we get the probability of the union in terms of values we know. And, as you indicated in the original query, the overlap that is counted twice among the two events does indeed need to be accounted for (and that is what it is subtracted from the sum).