Two events are independent when knowing one happened does not change the chance of the other: $P(A \cap B) = P(A)P(B)$, as in independent events. For three or more events, and for events that cannot happen together, the details matter.
Key idea
$A$, $B$, $C$ are (mutually) independent when all four equations hold: $P(A \cap B) = P(A)P(B)$, $P(A \cap C) = P(A)P(C)$, $P(B \cap C) = P(B)P(C)$ and $P(A \cap B \cap C) = P(A)P(B)P(C)$. Disjoint events with positive probabilities are never independent.
Why it works
Independence means information about some of the events tells you nothing about the others. If A and B are disjoint and $P(A) > 0$, then learning A happened tells you B did not happen: $P(B \mid A) = 0 \ne P(B)$. Using the definition of conditioning, $P(A \cap B) = 0 < P(A)P(B)$. Pairs are not enough for three events: the triple equation says that knowing both A and B tells you nothing about C, and that is a separate fact.
Worked examples
Worked example: Disjoint is dependent
A die is rolled. A = "roll a 1", B = "roll a 6". Are A and B independent?
- $P(A \cap B) = 0$, since one roll cannot be both.
- $P(A)P(B) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \ne 0$.
Answer: No. They are mutually exclusive, which makes them dependent.
Worked example: At least one of three
Three independent alarms sound in a fire with probabilities $0.9$, $0.8$ and $0.7$. Find the probability that at least one sounds.
- Complement: all three fail. By independence, $$P(\text{none}) = 0.1 \times 0.2 \times 0.3 = 0.006$$
- $$P(\text{at least one}) = 1 - 0.006 = 0.994$$
- For "exactly one", add three disjoint cases: $$0.9(0.2)(0.3) + 0.1(0.8)(0.3) + 0.1(0.2)(0.7) = 0.054 + 0.024 + 0.014 = 0.092$$
Answer: $0.994$ (and $0.092$ for exactly one)
Worked example: Independent in pairs only
Flip two fair coins. A = first is heads, B = second is heads, C = the flips match. Check pairwise and mutual independence.
- Each event has probability $\frac{1}{2}$. $A \cap B = \{HH\}$, $A \cap C = \{HH\}$, $B \cap C = \{HH\}$: each has probability $\frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2}$. So every pair is independent.
- $A \cap B \cap C = \{HH\}$ has probability $\frac{1}{4}$, but $P(A)P(B)P(C) = \frac{1}{8}$.
Answer: Independent in pairs, but not mutually independent: knowing A and B forces C.
Common mistake
Treating mutually exclusive events as independent ("they have nothing to do with each other"). Exclusive events are strongly linked: one rules out the other. The second slip is checking only pairs of events. Check: for three events write down all four products, including the triple one.