Independence and Dependence of Events
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Direct answer
Independence is a multiplicative fact: events A and B are independent exactly when P(A ∩ B) = P(A)P(B), equivalently P(A | B) = P(A) whenever P(B) > 0 — knowing B occurred does not reshape the odds of A. Pairwise independence among three or more events does not upgrade to mutual independence, which requires the product rule for every subcollection; the standard counterexample uses three events built from two coin tosses that pass every pairwise test yet fail the triple. Mutually exclusive events with positive probabilities are the opposite of independent: P(A ∩ B) = 0 can never equal P(A)P(B) > 0, so disjoint events are strongly dependent. For random variables, zero correlation (Cov = 0) is necessary but not sufficient for independence — an asymmetry JEE tests through true/false statements and joint-distribution checks.
What you must remember
- Definition: A, B independent ⟺ P(A ∩ B) = P(A)P(B); check this one product, nothing else.
- Complement stability: if A and B are independent, so are A and B', A' and B, A' and B' — the pair most used in "at least one" questions via P(A ∪ B) = 1 - P(A')P(B').
- Disjoint means dependent: for P(A), P(B) > 0, mutual exclusivity forces P(A ∩ B) = 0 ≠ P(A)P(B); independence and exclusivity coexist only when one event has probability zero.
- Pairwise versus mutual: pairwise products matching is not enough; mutual independence demands P(A ∩ B ∩ C) = P(A)P(B)P(C) as well, plus every pairwise condition.
- Coin-toss counterexample: with two fair coins, E1 = head on first, E2 = head on second, E3 = exactly one head: every pair is independent (each intersection has probability 1/4), yet P(E1 ∩ E2 ∩ E3) = 0 ≠ 1/8.
- Independent trials: Bernoulli structure — n independent repetitions with constant p — is where binomial probabilities come from; sampling with replacement preserves independence, without replacement destroys it.
- Zero correlation ≠ independence: for variables, Cov(X, Y) = 0 follows from independence but does not imply it; the classic counterexample is X uniform on {-1, 0, 1} with Y = X^2, uncorrelated yet functionally tied.
Building the counterexample
Toss two fair coins, sample space {HH, HT, TH, TT} with each outcome 1/4. Define E1 = head on the first coin = {HH, HT}, E2 = head on the second = {HH, TH}, and E3 = exactly one head = {HT, TH}. Each event has probability 1/2. Check pairwise: P(E1 ∩ E2) = P(HH) = 1/4 = (1/2)(1/2); P(E1 ∩ E3) = P(HT) = 1/4 = product again; P(E2 ∩ E3) = P(TH) = 1/4 likewise. All three pairs pass. Now the triple: E1 ∩ E2 ∩ E3 demands HH and exactly one head simultaneously — impossible, probability 0, while P(E1)P(E2)P(E3) = 1/8. Mutual independence fails at exactly one equation. This one construction answers a remarkable range of exam questions: it certifies that pairwise does not imply mutual, it supplies a dependent triple whose events look unrelated, and it shows independence is a property of the full collection, not of any pair in isolation. Alongside it, keep the simplest dependent pair: drawing two aives without replacement — first ace drawn changes the second ace probability from 4/52 to 3/51, so the events are dependent, whereas replacement restores independence.
How the exam frames it
JEE Main tests the definition directly: given P(A), P(B), P(A ∪ B) or P(A | B), decide independence, or compute P(A ∩ B) under the independence assumption in machine-fires and component-fails setups — the phrase "independently" in the question stem is the license to multiply. Advanced builds the deeper items: verify pairwise-but-not-mutual independence on a die sample space, count independent pairs among derived events, or test independence of functions of a random variable. The dependable traps: conflating mutually exclusive with independent (options always include both); assuming P(A | B) = P(B | A) — the confusion of directions; and inferring independence of X and Y from zero covariance. A derived-event subtlety worth internalising: if A and B are independent, then A and B' are too — but events built from the same trial, like A and A itself, or A and its subsets, are dependent. The topic anchors the probability unit of the JEE Main syllabus and recurs inside conditional-probability and Bayes questions in Advanced.
Frequently asked questions
What is the defining condition for two events to be independent?
P(A ∩ B) = P(A)P(B) — equivalently P(A | B) = P(A) when P(B) > 0; occurrence of one leaves the other's probability untouched.
Can two mutually exclusive events ever be independent?
Only if one has probability zero; otherwise P(A ∩ B) = 0 cannot equal the positive product P(A)P(B), so disjoint events are dependent.
How does pairwise independence differ from mutual independence?
Pairwise requires the product rule for every pair; mutual additionally requires P(A ∩ B ∩ C) = P(A)P(B)P(C) — the two-coin/three-event counterexample passes all pairwise tests but fails the triple.
Why does sampling without replacement destroy independence?
Because each draw changes the composition for the next: after an ace, the chance of another ace drops from 4/52 to 3/51, so the events of successive draws are dependent.
Does zero correlation between random variables imply independence?
No — zero covariance is necessary but not sufficient; X uniform on {-1, 0, 1} and Y = X^2 are uncorrelated yet completely determined by each other.