The only two rules you really need
Almost every basic probability question is the addition rule or the multiplication rule in a costume. The addition rule: P(A or B) = P(A) + P(B) − P(A and B): the overlap subtracted because it was counted twice. The multiplication rule: P(A and B) = P(A) × P(B | A), and when the events are independent, P(B | A) is just P(B), so it collapses to a plain product.
P(A) = 0.4, P(B) = 0.5, independent. Both: 0.4 × 0.5 = 0.20. Either: 0.4 + 0.5 − 0.20 = 0.70. Neither: 1 − 0.70 = 0.30. And P(A | B) = 0.4, exactly P(A), because independence means B's happening carries no news about A. Make the same events mutually exclusive instead and the picture changes completely: both becomes 0, either becomes 0.90. Which relationship holds is the whole question: the arithmetic is the easy part.
The three relationships, kept straight
- Independent: one happening does not move the other's probability (separate coin flips, separate draws with replacement). Intersection = product.
- Mutually exclusive: they cannot co-occur at all (one die roll showing both 3 and 5). Intersection = 0, so conditional probabilities are 0 too. Exclusive events are never independent (unless one is impossible): learning B happened destroys A entirely.
- Neither: you need the intersection from data or a model; then the conditionals fall out of their definition.
Adding P(A) + P(B) for "either" and getting something over 1: the signature of a forgotten intersection. Probabilities of unions can never exceed 1; if yours does, the overlap is still in there twice.
Frequently asked questions
What is P(A or B) when the events are independent?
P(A) + P(B) − P(A)·P(B). Independence simplifies the intersection, not the union: the subtraction still applies.
Can P(A|B) be bigger than P(A)?
Yes, that is precisely what 'B is evidence for A' means. It can also be smaller, or (under independence) exactly equal.