Probability: What is the mathematical meaning of "or" in this sentence?

Question

I'm having trouble understanding the description of the event in italics (with P = 0.8) in the question below.

"The probability that an American industry will locate in Shanghai is P(S) = 0.7.
The probability that it will locate in Beijing is P(B) = 0.4.
And the probability that it will locate in either Shanghai or Beijing or both is 0.8.
What is the probability that the industry will locate: (a) in both cities? (b) in neither city?"

Doubt: Why the answer to (a) isn't P = 0.8?

The question itself says that "the probability that it will locate in either Shanghai or Beijing or both is 0.8". Isn't it saying P(A∪B) = P(A∩B) = 0.8?

I've found many solutions available and the answers are (a) 0.3 and (b) 0.2 but I still cannot understand why it is not 0.8.

Thanks.

Answer 1

Recall that the union of sets $A$ and $B$ consists of things that belong to either $A$ or $B$ or both. This is exactly the wording in your problem. By contrast, “$A$ or $B$, but not both” is an exclusive “or,” so you have to remove everything that’s in both $A$ and $B$, i.e., it corresponds to $(A\cup B)\setminus(A\cap B)$.

This is also why you have the inclusion/exclusion formula $\Pr(A\cup B) = \Pr(A)+\Pr(B)-\Pr(A\cap B)$. If you simply add the probabilities of $A$ and $B$ together, you double-count all of the elements they have in common: they’ve already been accounted for in $\Pr(A)$

Answer 2

It's actually saying $P(A \cup B) = .8$; from here, the probability that it will locate in both is $P(A \cap B)$ since it is asking that both event $A$ and event $B$ hold. To get $P(A \cap B)$, utilize the formula:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B).$$