Decide when two or three events are independent, use $P(A \cap B) = P(A)P(B)$, tell independence from mutual exclusivity, and handle "at least one" with independent trials.
If the result of a coin toss tells you nothing about a die thrown afterwards, the two are independent. In symbols, $A$ and $B$ are independent when knowing that $A$ occurred leaves the probability of $B$ unchanged: $$P(B \mid A) = P(B).$$ Put this into the multiplication rule $P(A \cap B) = P(A)P(B \mid A)$ and the usual working form appears: $$A \text{ and } B \text{ are independent} \iff P(A \cap B) = P(A),P(B).$$ To test independence, calculate $P(A)$, $P(B)$ and $P(A \cap B)$ and compare.
Example: $P(A) = 0.3$, $P(B) = 0.5$ and $P(A \cup B) = 0.65$. Are $A$ and $B$ independent? By the addition rule, $P(A \cap B) = 0.3 + 0.5 - 0.65 = 0.15$. Now $P(A)P(B) = 0.3 \times 0.5 = 0.15$. The two are equal, so $A$ and $B$ are independent.
Example: A fair die is thrown. $A$ = "even" and $C$ = "greater than $4$". $P(A) = \dfrac{1}{2}$, $P(C) = \dfrac{1}{3}$ and $A \cap C = {6}$ so $P(A \cap C) = \dfrac{1}{6} = \dfrac{1}{2} \times \dfrac{1}{3}$: independent. But for $D$ = "greater than $3$": $P(D) = \dfrac{1}{2}$, $A \cap D = {4, 6}$ gives $\dfrac{1}{3} \ne \dfrac{1}{2} \times \dfrac{1}{2}$, so $A$ and $D$ are not independent.
If $A$ and $B$ are independent then so are $A$ and $B'$. Proof: $P(A \cap B') = P(A) - P(A \cap B) = P(A) - P(A)P(B) = P(A)\bigl(1 - P(B)\bigr) = P(A)P(B')$. Applying this twice shows $A'$ and $B'$ are independent too. This is why we can multiply "the probability that $A$ does not happen" into a product without checking again.
These are different ideas and are often confused.
| Mutually exclusive | Independent | |
|---|---|---|
| Meaning | cannot occur together | one does not affect the other |
| Test | $P(A \cap B) = 0$ | $P(A \cap B) = P(A)P(B)$ |
| Addition / multiplication | $P(A \cup B) = P(A) + P(B)$ | $P(A \cap B) = P(A)P(B)$ |
If $P(A) > 0$ and $P(B) > 0$ and the events are mutually exclusive, then $P(A \cap B) = 0$ but $P(A)P(B) > 0$, so they are not independent: if one occurs, the other becomes impossible.
$A$, $B$, $C$ are pairwise independent if $$P(A \cap B) = P(A)P(B), \quad P(B \cap C) = P(B)P(C), \quad P(A \cap C) = P(A)P(C).$$ They are mutually independent if, in addition, $P(A \cap B \cap C) = P(A)P(B)P(C)$. Pairwise independence is weaker: toss a fair coin twice and let $A$ = "first is a head", $B$ = "second is a head" and $C$ = "both tosses agree". Each pair is independent (every pair has probability $\dfrac{1}{4} = \dfrac{1}{2} \times \dfrac{1}{2}$), but $P(A \cap B \cap C) = \dfrac{1}{4}$ whereas $P(A)P(B)P(C) = \dfrac{1}{8}$.
If $A_1, \dots, A_n$ are independent with probabilities $p_1, \dots, p_n$, the probability that none occurs is $(1 - p_1) \cdots (1 - p_n)$, so $$P(\text{at least one}) = 1 - (1 - p_1)(1 - p_2) \cdots (1 - p_n).$$
Example: Three marksmen hit a target with probabilities $0.6$, $0.7$ and $0.8$, firing independently once each. Find the probability that (i) at least one hits, (ii) exactly one hits. (i) $1 - (0.4)(0.3)(0.2) = 1 - 0.024 = 0.976$. (ii) Exactly one hits: $0.6 \times 0.3 \times 0.2 + 0.4 \times 0.7 \times 0.2 + 0.4 \times 0.3 \times 0.8 = 0.036 + 0.056 + 0.096 = 0.188$.
Exam tip: Examiners like "show that $A$ and $B$ are independent" and "are $A$ and $B$ mutually exclusive?" in the same question. Answer by calculation: for independence show $P(A \cap B) = P(A)P(B)$; for mutual exclusivity show $P(A \cap B) = 0$. Remember the link: two events with positive probabilities cannot be both. Also, when the question says the trials are "independent", you may multiply without using conditional probability.