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3 Responses
Great question-there’s a common mix-up between “independent” and “mutually exclusive.” Independence means knowing B doesn’t change the chance of A, i.e., P(A|B) = P(A); it does not mean you add probabilities. In fact, the rule for “and” is P(A and B) = P(A)×P(B) when A and B are independent. You can see this two ways. Sample-space view: flipping a coin and rolling a die gives 12 equally likely outcomes, and only one is (H,6), so P(H and 6) = 1/12 = (1/2)×(1/6). Chain-rule view: P(A∩B) = P(A|B)P(B); with independence P(A|B) = P(A), so P(A∩B) = P(A)P(B). Addition, on the other hand, is for “or”: P(A or B) = P(A) + P(B) − P(A and B). If events are mutually exclusive (can’t occur together), then P(A and B)=0 and the “or” becomes a simple sum. But H and 6 are not mutually exclusive-they can happen together-so adding for “and” would be wrong; it would give 1/2 + 1/6 = 2/3, which is clearly too big for something rarer than either event alone. Does it help to contrast with a truly exclusive pair, like “H” and “T” on the same coin toss, where P(H and T) = 0 but P(H or T) = 1/2 + 1/2 = 1?
For independent events, “and” means multiply: P(heads and 6) = P(heads)P(6) = 1/2 · 1/6 = 1/12, while adding is for “or” (and equals P(A)+P(B) only when the events are mutually exclusive-basically independent in a different sense). When I first learned this, I drew a simple 2×6 grid and saw only one H–6 cell out of 12; this Khan Academy explainer is clear: https://www.khanacademy.org/math/statistics-probability/probability-library/probability-independent-events/v/probability-with-independent-events
“Two strangers on a bus” is a great image! Independence means they don’t change each other’s chances, not that we add them. The general rule is P(A and B) = P(A)·P(B) because “both” is like first A happens, then given that, B still has its usual chance: P(A and B) = P(A)·P(B|A), and independence says P(B|A) = P(B). For the coin and die: P(heads) = 1/2, P(6) = 1/6, so P(heads and 6) = (1/2)(1/6) = 1/12. You can also see it by counting: there are 12 equally likely outcomes (H/T times 1–6), and only one is (H,6). Adding is for “or”: P(A or B) = P(A) + P(B) − P(A and B); if events are mutually exclusive (can’t both happen), then the overlap is zero and addition works-but independence is not the same as mutual exclusivity. In fact, independent events usually do overlap! About your note: for independence, P(A|B) = P(A) (exactly equal), while P(A|B) > P(A) means B actually makes A more likely, so they’re positively associated, not independent. Does it help to sketch the 2×6 grid and then try computing P(heads or 6) as a quick check (you should get 7/12)?