Independent events

One card is drawn from a pack of 52. E: it is a spade, P(E) = 1/4; F: it is an ace, P(F) = 1/13; E ∩ F: the ace of spades, 1/52. Then P(E|F) = 1/4 = P(E) and P(F|E) = 1/13 = P(F): neither event changes the chance of the other.

E and F are independent if P(F|E) = P(F) (with P(E) ≠ 0) and P(E|F) = P(E) (with P(F) ≠ 0).

Equivalent working definition: E and F are independent if P(E ∩ F) = P(E) P(F). If this fails, they are dependent.

Independent is not the same as mutually exclusive. Mutually exclusive events share no outcome, so P(E ∩ F) = 0. Two events with non-zero probabilities cannot be both: independent ones must overlap, and mutually exclusive ones are dependent.

Two experiments are independent if, for every event E of the first and F of the second, P(E ∩ F) = P(E) P(F).

Three events A, B, C are mutually independent only if four conditions hold. Each pair must multiply, P(A ∩ B) = P(A)P(B) and likewise for A with C and for B with C, and P(A ∩ B ∩ C) must equal the product P(A)P(B)P(C).

One die: E: a multiple of 3 = {3, 6}, F: even = {2, 4, 6}. P(E) = 1/3, P(F) = 1/2, P(E ∩ F) = 1/6 = (1/3)(1/2), so E and F are independent. Two throws: 'odd on the first' and 'odd on the second' are independent, 1/4 = (1/2)(1/2).

Three coins. E: three heads or three tails (1/4), F: at least two heads (1/2), G: at most two heads (7/8). P(E ∩ F) = 1/8 = P(E)P(F): independent. P(E ∩ G) = 1/8 ≠ 7/32 and P(F ∩ G) = 3/8 ≠ 7/16: both pairs dependent.

If E and F are independent, so are E and F′, E′ and F, and E′ and F′. For instance P(E ∩ F′) = P(E) − P(E)P(F) = P(E)P(F′).

For independent A and B, P(at least one occurs) = P(A ∪ B) = 1 − P(A′)P(B′).

A and B throw a die in turn until a six appears; A starts. A can win only on throw 1, 3, 5, …, with probabilities 1/6, (25/36)(1/6), (25/36)²(1/6), …. This geometric series sums to (1/6)/(1 − 25/36) = 6/11, so P(A wins) = 6/11 and P(B wins) = 5/11.