Working with independent events

If E and F are independent, then so are E and F′, E′ and F, and E′ and F′. So the chance that neither happens is P(E′) × P(F′).

This gives the most useful trick in the chapter:

P(at least one) = 1 – P(none)

Example: three people shoot at a target, hitting with chances 1/2, 1/3 and 1/4, independently. P(all miss) = 1/2 × 2/3 × 3/4 = 1/4. So P(target is hit) = 1 – 1/4 = 3/4. Far quicker than adding up all the ways to hit.

Exactly one of A and B: P(A)P(B′) + P(A′)P(B), either A without B, or B without A. With P(A) = 0.5 and P(B) = 0.2: 0.5 × 0.8 + 0.5 × 0.2 = 0.5.

Taking turns: A and B toss a coin in turn, A first, and the first head wins. A wins on toss 1, or after two tails, or after four tails, and so on:

P(A wins) = 1/2 + (1/2)²(1/2) + (1/2)⁴(1/2) + ... This is a geometric series, and its sum is a/(1 – r) = (1/2)/(1 – 1/4) = 2/3.

Check consistency of given numbers: any probability must lie between 0 and 1, and P(A ∩ B) can never be bigger than P(A) or P(B).