L3 · Non-Mutually Exclusive Events: The Addition Rule

Learning goal

“I can compute P(A ∪ B) with the addition rule - adding the probabilities and, when the events overlap, subtracting the double-counted intersection - and I can decide which form of the rule a problem needs.”

Addition rule (mutually exclusive form) — if A and B cannot occur together, P(A ∪ B) = P(A) + P(B).

Addition rule (general form) — for any two events, P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

Double-counted — what happens to the overlap when P(A) and P(B) are added naively: outcomes in both events are counted twice, so one copy must be subtracted.

Worked example

One die roll. Let A = roll a 1 and B = roll a 6. Step 1: Check the intersection: a single roll cannot show both 1 and 6, so A ∩ B = ∅ - the events are mutually exclusive. Step 2: Apply the short rule: P(A ∪ B) = P(A) + P(B) = 1/6 + 1/6 = 2/6 = 1/3. Step 3: Verify by counting: A ∪ B = {1, 6} - 2 faces out of 6. Say it plainly: when the overlap is empty, there is nothing to double-count, and adding is exact.

Check yourself — explain it

In the mutually exclusive form, why is no subtraction term needed?

Paired problem — example still visible

One die roll: A = roll a 3, B = roll a 4. Find P(A ∪ B). -> The events are mutually exclusive (one roll, one face): P = 1/6 + 1/6 = 1/3. Counting check: {3, 4} = 2 faces

Faded completion — last step is yours

"A bag holds 5 red, 3 blue and 2 green marbles. One marble is drawn; A = red and B = blue are mutually exclusive, since one marble has one colour. P(A ∪ B) = 5/10 + ___ = ___." -> 3/10; 8/10 = 4/5

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