Does P(A or B) just mean P(A) + P(B)?

Here is a question that tests this misconception head-on, in the spirit of JEE Advanced 2021 Paper 2 Q17:

The misconception answer adds the two counts: 33 multiples of 3, 14 multiples of 7, so $\tfrac{33}{100} + \tfrac{14}{100} = \tfrac{47}{100}$. But the multiples of $21$ — namely $21, 42, 63, 84$ — sit in both lists, so each has been counted twice. The correct move subtracts that overlap once:

The answer is $\tfrac{43}{100}$, not $\tfrac{47}{100}$ — smaller by exactly the four double-counted multiples of 21.

"Or" reads as "add" in plain English, and for sets that happen not to overlap the additive rule even gives the right answer, so it survives unchallenged for a long time. Because nothing in the rule caps the total, two large overlapping sets can push the "probability" above 1 — a result the student often does not notice they have produced.

Garfield and Ahlgren, and Fischbein, document this additivity-overextension error: students extend the disjoint-additivity axiom $P(A \cup B) = P(A) + P(B)$ to events that are not disjoint. Being shown the inclusion–exclusion formula rarely dislodges it, because adding still feels like what "or" means. What dislodges it is handing the student one specific number — $21$ — and letting them watch it sit in both counts, so the double-counting becomes a thing they see rather than a principle they are told.

Include both sets by adding them, then exclude the overlap you counted twice by subtracting it once: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. The $-\,P(A \cap B)$ is exactly the part that lives in both, removed the one extra time it was tallied.

Adding alone is correct only when the events are mutually exclusive, that is when $P(A \cap B) = 0$ and they cannot both occur. For three events the principle extends with alternating signs,

Count the numbers from $1$ to $100$ that are a multiple of 3 or 7, watching the overlap.

1. Multiples of 3. $\lfloor 100/3 \rfloor = 33$ of them.
2. Multiples of 7. $\lfloor 100/7 \rfloor = 14$ of them.
3. The overlap. A number in both lists is a multiple of $3 \times 7 = 21$. From 1 to 100 these are $21, 42, 63, 84$ — $\lfloor 100/21 \rfloor = 4$ of them. Each was counted once among the 33 and again among the 14.
4. Subtract the overlap once.
   $$33 + 14 - 4 = 43 \quad\Rightarrow\quad P = \tfrac{43}{100}.$$
5. Compare. The additive rule gave $\tfrac{47}{100}$; the correct answer is $\tfrac{43}{100}$, smaller by exactly $\tfrac{4}{100}$ — the four multiples of 21 that were tallied twice. The error is the size of the overlap, nothing mysterious.

The same framework settles JEE Advanced 2022 Paper 1 Q3, which asks for $P(\text{at most one symptom})$: this is the complement of a union, and unpacking the union needs inclusion–exclusion so that students sharing two symptoms are not counted twice. And on a class with $P(\text{football}) = 0.5$, $P(\text{tennis}) = 0.4$, $P(\text{both}) = 0.2$, the chance of football or tennis is $0.5 + 0.4 - 0.2 = 0.7$, not the additive $0.9$.