In short: P(A or B) = P(A) + P(B) − P(A∩B). You subtract the intersection because outcomes that satisfy both A and B get counted once in P(A) and again in P(B) — the subtraction removes that double-count. Only when A and B are mutually exclusive (P(A∩B) = 0) can you simply add.
A huge share of marks lost in JEE Advanced probability come from one reflex: writing P(A or B) = P(A) + P(B). That is the reason you subtract the intersection in a probability union — without it you count the overlap twice. The inclusion-exclusion principle is not an optional refinement; it is the correct definition of a union's probability. Get comfortable with it and a whole class of "or" problems becomes mechanical.
The reliable method
For any two events A and B, use inclusion-exclusion:
- Spot the "or". Words like "or", "either", "at least one of A, B", or "A union B" signal a union.
- Write the full formula. P(A∪B) = P(A) + P(B) − P(A∩B). Never drop the last term by default.
- Find the overlap P(A∩B). Identify the outcomes in both events. If A and B are independent, P(A∩B) = P(A)·P(B); otherwise count them directly.
- Subtract once. The overlap was added twice (once in P(A), once in P(B)), so subtracting it once leaves it counted exactly once.
- Only skip the subtraction when truly disjoint. If A and B cannot both occur, P(A∩B) = 0 and the formula collapses to P(A) + P(B).
A worked example
One card is drawn at random from a standard 52-card deck. Let A be "the card is a heart" and B be "the card is a face card" (Jack, Queen, or King). Find P(A or B).
Count each piece:
- P(A) = 13/52 (thirteen hearts)
- P(B) = 12/52 (four suits × three face cards)
- P(A∩B) = 3/52 (the heart Jack, heart Queen, heart King)
If you naively add, you get 13/52 + 12/52 = 25/52 — but that counts the three heart face cards twice, once as hearts and once as face cards. Apply inclusion-exclusion:
P(A∪B) = P(A) + P(B) − P(A∩B) = 13/52 + 12/52 − 3/52 = 22/52 = 11/26 ≈ 0.423
So the correct probability is 11/26, not 25/52. The method holds because the three cards in A∩B are members of both sets; the formula adds them in twice and the single subtraction restores them to a count of one each.
Common mistakes to avoid
- Writing P(A∪B) = P(A) + P(B) by default. Fix: always include the − P(A∩B) term unless you have proven the events are disjoint.
- Assuming "different categories" means disjoint. Fix: hearts and face cards are different categories but still overlap — check for shared outcomes explicitly.
- Confusing mutually exclusive with independent. Fix: mutually exclusive means P(A∩B) = 0; independent means P(A∩B) = P(A)·P(B). They are different conditions.
- Computing the overlap wrong. Fix: for independent events use P(A)·P(B); for everything else, count the actual outcomes in both sets.
- Forgetting to extend to three sets. Fix: P(A∪B∪C) adds the singles, subtracts the pairwise overlaps, then adds back P(A∩B∩C).
Frequently asked questions
Why do you subtract P(A and B) when finding P(A or B)? Because outcomes in both A and B are counted once inside P(A) and again inside P(B). Subtracting P(A∩B) once removes the duplicate so each outcome is counted exactly once.
When can I just add P(A) and P(B)? Only when A and B are mutually exclusive, meaning they cannot happen together and P(A∩B) = 0. Then the subtraction term vanishes.
What is the inclusion-exclusion formula for three events? P(A∪B∪C) = P(A)+P(B)+P(C) − P(A∩B) − P(A∩C) − P(B∩C) + P(A∩B∩C). You alternate subtracting and adding overlaps.
Does mutually exclusive mean the same as independent? No. Mutually exclusive events have zero overlap; independent events satisfy P(A∩B) = P(A)·P(B). Two events that are mutually exclusive with non-zero probabilities are in fact dependent.
How do I find P(A∩B) if it is not given? If A and B are independent, multiply: P(A∩B) = P(A)·P(B). Otherwise count the outcomes that lie in both events directly.
Practise this
Find out if this trips you up — [take the free diagnostic](/diagnostic), then work through the [free Socratic lesson](/lesson/start) to make inclusion-exclusion automatic.