You want the probability that one thing or another happens. The obvious move is to add the two probabilities together — and the obvious move is wrong, because any outcome that satisfies both events has just been counted twice, once in each total. The fix is a single subtraction: P(A or B) = P(A) + P(B) − P(A and B). Drag the overlap below and the Venn diagram redraws itself with areas that are genuinely proportional to the probabilities, the sum is taken apart to expose the double count, and a real simulation runs to see which answer the world actually agrees with.
The two events, drawn to scale
Why the subtraction is there
A run of real draws
running proportion so farthe rule’s answerjust adding themordinary sampling wobble
The addition ruledrag the overlap and watch
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Send the overlap to
200
P(A)——
P(B)——
P(A and B)——
P(A or B)——
Way of answering
P(A or B)
Off by
Just add them
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Addition rule
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This run
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The middle row is the truth for this sample space — it is arithmetic on outcomes, not an estimate, so it has no error to report. The top row is what you get by forgetting the subtraction. The bottom row is what a finite run of draws actually delivered, and it should hover around the middle row and never around the top one.
The whole rule is a counting idea wearing probability clothing, and it is worth holding it in that plainer form. Suppose two lists have been handed to you: everyone who owns a car, and everyone who owns a bicycle. You want to know how many distinct people appear on at least one list. If you simply add the lengths of the two lists you have not counted the people, you have counted the list entries — and anyone who owns both appears as two entries while remaining one person. Striking out one of each duplicate pair is the entire correction, and P(A and B) is precisely the number of duplicate pairs. Everything else follows from that single observation. When the lists share nobody, there is nothing to strike out and adding works unchanged — which is why “mutually exclusive” is not a different rule but the same rule with a zero in it. And when the two lists are long enough that their lengths sum past the size of the whole group, you have proved there are duplicates without inspecting a single name.
Five ways this goes wrong. First, “or” here is inclusive. P(A or B) counts outcomes where A happens, where B happens, and where both happen together. If you actually want exactly one of the two and not both, that is a different quantity — P(A) + P(B) − 2 P(A and B) — because the shared outcomes must be removed twice, not once, to take them out altogether. Second, mutually exclusive is not the same as independent, and this is the single most common confusion in the chapter. Mutually exclusive means the two events cannot both occur; independent means learning that one occurred tells you nothing about the other. They are close to opposites: if two events with non-zero probability are mutually exclusive, then knowing A happened tells you B definitely did not, which is about as far from “tells you nothing” as information gets. Third, a sum above 1 is not a probability, it is a receipt for a mistake. Nothing can happen more often than always, so any time adding gives you more than 100% you have proved the events overlap — and by at least the amount of the excess. Fourth, you cannot get the overlap out of P(A) and P(B). Two events with the same individual probabilities can overlap completely, not at all, or anywhere in between, and the diagram beside you will show you all three on demand. The joint probability is separate information: you either count it, or you are given it, or you assume something (like independence) that lets you compute it. Fifth, subtracting when there is no overlap costs nothing. Some students learn two rules and then agonise over which applies. There is only one rule; when the events are disjoint the term you subtract is zero and it quietly disappears.