Why does P(A or B) = P(A) + P(B) give the wrong answer when A and B overlap?

## The rule For any two events `A` and `B` in the same sample space, `P(A ∪ B) = P(A) + P(B) - P(A ∩ B)` This is the **general addition rule**, also called two-set inclusion-exclusion. The subtracted term is the whole point: outcomes belonging to both events were counted once inside `P(A)` and again inside `P(B)`, so they must be removed once to leave them counted exactly once. ## Where it comes from The rule is not an extra assumption; it follows from the additivity axiom, which says probabilities of **disjoint** events add. Split the union into two disjoint pieces: `A ∪ B = A ∪ (B minus A)` These two pieces share no outcome, so `P(A ∪ B) = P(A) + P(B \ A)`. Now split `B` the same way: `B = (A ∩ B) ∪ (B \ A)`, again disjoint, so `P(B) = P(A ∩ B) + P(B \ A)`, which rearranges to `P(B \ A) = P(B) - P(A ∩ B)`. Substituting gives `P(A ∪ B) = P(A) + P(B) - P(A ∩ B)`. Every step used only additivity over disjoint sets, which is why the rule holds for *any* two events, however they are related. ## The worked example Draw one card at random from a standard 52-card deck. All 52 outcomes are equally likely. - `R` = the card is red: 26 outcomes, `P(R) = 26/52 = 1/2`. - `F` = the card is a face card, meaning jack, queen or king: three per suit across four suits, so 12 outcomes, `P(F) = 12/52 = 3/13`. - `R ∩ F` = the card is a red face card: jack, queen and king of hearts plus the same three of diamonds, so 6 outcomes, `P(R ∩ F) = 6/52`. Apply the rule: `P(R ∪ F) = 26/52 + 12/52 - 6/52 = 32/52 = 8/13 ≈ 0.615` The naive sum `38/52 ≈ 0.731` is wrong by exactly `6/52`, the size of the overlap. You can confirm the correct value by direct counting: 26 red cards plus the 6 black face cards that are not already counted equals 32. ## The special case When `A ∩ B` is empty, the events are **mutually exclusive** (also called disjoint): they cannot both occur on the same trial. Then `P(A ∩ B) = 0` and the rule collapses to the simple form `P(A ∪ B) = P(A) + P(B)`. That is the only situation in which plain addition is exact. The most frequent error here is confusing mutual exclusivity with **independence**. They are different, in fact opposed: if `A` and `B` are mutually exclusive and both have positive probability, then learning that `A` occurred tells you `B` definitely did not, so they are strongly dependent. ## Companion results Several other rules fall out of the same additivity idea and are worth having ready: - **Complement rule**: `P(A^c) = 1 - P(A)`, because `A` and its complement are disjoint and together fill the sample space. When "at least one" is awkward to count directly, computing the complement is often far easier. - **Monotonicity**: if every outcome in `A` is also in `B`, then `P(A) ≤ P(B)`. - **Union bound**: dropping the subtracted term can only inflate the result, so `P(A ∪ B) ≤ P(A) + P(B)` always. This upper bound holds even when you know nothing about the overlap. - **Lower bound on the overlap**: since `P(A ∪ B) ≤ 1`, rearranging gives `P(A ∩ B) ≥ P(A) + P(B) - 1`. With `P(A) = 0.6` and `P(B) = 0.5`, the two events must overlap on at least 0.1 of the probability, no matter how they are arranged. ## Interview framing Interviewers like this question because the failure mode is quiet. A candidate who adds two marketing segment rates and reports 73% reach rather than 62% has produced a number that looks fine on a slide. Say the rule, name the overlap term, and state explicitly that the simple version needs disjointness. If you are given only the two marginal probabilities and no overlap, say so and give the bounds rather than inventing a joint value. ## Traps to avoid - Reading "or" as exclusive. In probability `A ∪ B` includes the outcomes where both events happen; if you truly want exactly one of them, that is `P(A) + P(B) - 2 P(A ∩ B)`. - Multiplying to get the overlap when the events are not independent. `P(A ∩ B) = P(A) P(B)` is a statement about independence, not a general identity. - Reporting a union probability above 1 and not noticing that this alone proves double counting.