When one event does not change the other: independent events and the test P(A ∩ B) = P(A) × P(B)
Two events are independent when knowing that one occurred does not change the probability of the other: P(A | B) = P(A). Putting this into the multiplication rule P(A ∩ B) = P(A | B) × P(B) gives the test: A and B are independent exactly when P(A ∩ B) = P(A) × P(B). So independence is settled by numbers, not by a feeling that two events are unrelated. A coin tossed together with a die is a clear case, because neither can influence the other: P(head and even) = 1/2 × 1/2 = 1/4, which matches 3 of the 12 pairs. But events on the same die can be tricky. Let A = “even” = {2, 4, 6}, B = “at most 4” = {1, 2, 3, 4} and C = “at most 3” = {1, 2, 3}. Then A ∩ B = {2, 4}, so P(A ∩ B) = 1/3 = 1/2 × 2/3, and A and B are independent. But A ∩ C = {2}, so P(A ∩ C) = 1/6 while P(A) × P(C) = 1/4, and A and C are dependent. The test, not intuition, decides.
| Events on one die | P(A ∩ B) | P(A) × P(B) | Verdict |
|---|---|---|---|
| A even, B at most 4 | 2/6 = 1/3 | 1/2 × 2/3 = 1/3 | independent |
| A even, C at most 3 | 1/6 | 1/2 × 1/2 = 1/4 | dependent |
| Situation | Independent? | Reason |
|---|---|---|
| a coin is tossed twice: head first, tail second | yes | the coin has no memory |
| a card is drawn, put back, and another is drawn | yes | the second draw sees the same cards |
| a card is drawn and kept, then another is drawn | no | one card fewer in the second draw |
| a pupil is late; the pupil’s bus is late | no | a late bus makes lateness likelier |
Write the 12 pairs of a coin and a die (H1, …, T6). Count n(S) = 12, n(head) = 6, n(even) = 6 and n(head and even) = 3. Check that n(A ∩ B) × n(S) = n(A) × n(B), that is 3 × 12 = 6 × 6, which is the test written with counts.