Difficulty: Intermediate | Prerequisites: Basic set theory, axioms of probability (Chapter 2 foundations).
This topic sits at the heart of probability theory: knowing whether two events influence each other changes how you calculate their joint and conditional probabilities. Independence is the bridge between simple single-event problems (Chapter 2 basics) and the multivariate reasoning you will need for statistical inference later in the course. If you can prove independence and test for it from given probabilities, you have the toolkit to tackle Bayes' theorem problems, joint distributions, and eventually hypothesis testing. You should already be comfortable with sample spaces, the addition rule, and the complement rule before working through this material.
Two events are independent when knowing one happened tells you nothing about whether the other happened. The formal test is P(A ∩ B) = P(A) · P(B). If that equality holds, they are independent; if not, they are dependent. This single definition drives every proof and calculation in this topic.
Independent events
Two events A and B such that the occurrence of one does not affect the probability of the other. Formally, P(A ∩ B) = P(A) · P(B).
Think of it as: learning that A happened gives you zero new information about B.
Dependent events
Two events where knowing one changes the probability of the other. P(A ∩ B) ≠ P(A) · P(B).
In simple terms, this means: if A happening makes B more or less likely, they are dependent.
Complement of an event (Ā or Aᶜ)
Everything in the sample space that is not A. P(Ā) = 1 − P(A).
Think of it as: the "opposite" outcome. If A is "rain," then Ā is "no rain."
Joint probability, P(A ∩ B)
The probability that both A and B occur together.
In simple terms, this means: what are the chances both things happen at once?
Union probability, P(A ∪ B)
The probability that at least one of A or B occurs.
Think of it as: the chance that A happens, or B happens, or both.
Conditional probability, P(A|B)
The probability of A given that B has already occurred. P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0.
In simple terms, this means: if you already know B happened, how likely is A now?
Claim: If A and B are independent, then A and B̄ are independent.
Strategy: Show that P(A ∩ B̄) = P(A) · P(B̄).
Proof outline:
Start from P(A) = P(A ∩ B) + P(A ∩ B̄), because B and B̄ partition the sample space.
Rearrange: P(A ∩ B̄) = P(A) − P(A ∩ B).
Since A and B are independent, substitute P(A ∩ B) = P(A) · P(B).
P(A ∩ B̄) = P(A) − P(A) · P(B) = P(A) · (1 − P(B)) = P(A) · P(B̄). ∎
Takeaway: Independence is "contagious" to complements. If you prove A ⊥ B, you automatically get A ⊥ B̄, Ā ⊥ B, and Ā ⊥ B̄.
Claim: If P(A) = 1, then A and B are independent for any event B.
Strategy: Show P(A ∩ B) = P(A) · P(B) = 1 · P(B) = P(B).
Proof outline:
Since A has probability 1, its complement Ā has P(Ā) = 0.
P(A ∩ B) = P(B) − P(Ā ∩ B). But P(Ā ∩ B) ≤ P(Ā) = 0, so P(Ā ∩ B) = 0.
Therefore P(A ∩ B) = P(B) = P(A) · P(B). ∎
Takeaway: A certain event (probability 1) is independent of everything, which makes intuitive sense: learning that a certainty occurred changes nothing.
The method is always the same: compute P(A ∩ B) from the information you have, then check whether it equals P(A) · P(B).
Using the addition rule: If you know P(A), P(B), and P(A ∪ B), recover the intersection: P(A ∩ B) = P(A) + P(B) − P(A ∪ B). Then compare to P(A) · P(B).
Example: P(A) = 0.5, P(B) = 0.5, P(A ∪ B) = 0.75. Then P(A ∩ B) = 0.5 + 0.5 − 0.75 = 0.25. And P(A) · P(B) = 0.25. Equal, so A and B are independent.
Using De Morgan's law: If you know P(Eᶜ ∩ Fᶜ), apply De Morgan: P(Eᶜ ∩ Fᶜ) = P((E ∪ F)ᶜ) = 1 − P(E ∪ F). Solve for P(E ∪ F), then recover P(E ∩ F) via the addition rule.
Example: P(E) = 0.4, P(F) = 0.5, P(Eᶜ ∩ Fᶜ) = 0.3. Then P(E ∪ F) = 1 − 0.3 = 0.7. So P(E ∩ F) = 0.4 + 0.5 − 0.7 = 0.2. And P(E) · P(F) = 0.20. Equal, so E and F are independent.
Using conditional probabilities: If you know P(I|J) and P(Iᶜ|Jᶜ), you need to recover P(I), P(J), and P(I ∩ J). Use the law of total probability and the definition of conditional probability to set up equations.
P(I|J) = P(I ∩ J) / P(J) gives you P(I ∩ J) in terms of P(J).
P(Iᶜ|Jᶜ) = P(Iᶜ ∩ Jᶜ) / P(Jᶜ). Use De Morgan and the complement to express P(Iᶜ ∩ Jᶜ) = 1 − P(I ∪ J) = 1 − [P(I) + P(J) − P(I ∩ J)].
With two equations and the law of total probability P(I) = P(I|J)·P(J) + P(I|Jᶜ)·P(Jᶜ), solve for the unknowns, then check the product rule.
P(A \cap B) = P(A) \cdot P(B) \quad \text{(definition of independence)}P(A \cup B) = P(A) + P(B) - P(A \cap B) \quad \text{(addition rule)}P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0 \quad \text{(conditional probability)}P(\bar{A}) = 1 - P(A) \quad \text{(complement rule)}P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B) \quad \text{(De Morgan's law)}P(A) = P(A|B) \cdot P(B) + P(A|B^c) \cdot P(B^c) \quad \text{(law of total probability)}Independence assumptions underpin most of classical statistics: when you draw a random sample, you treat each observation as independent to justify formulas for standard errors and confidence intervals. In engineering, independence lets you multiply component reliabilities to get system reliability (if each part fails independently with probability p, a chain of n parts survives with probability (1 − p)ⁿ).
Students often think "independent" and "mutually exclusive" mean the same thing. They do not. Mutually exclusive events (A ∩ B = ∅) with nonzero probabilities are always dependent, because if one happens, the other cannot.
Students sometimes assume that if P(A|B) = P(A), they do not also need to verify P(A ∩ B) = P(A) · P(B). In fact, these two conditions are equivalent (given P(B) > 0), but the product rule is the definition; the conditional form is a consequence.
A common error is testing independence by comparing P(A) and P(B) to each other. Their individual sizes tell you nothing; only the relationship between the joint probability and the product of the marginals matters.
Students forget that independence of A and B automatically implies independence of A and B̄, Ā and B, and Ā and B̄. You do not need to verify all four pairs separately.
⚠️ Proofs involving complements (A ⊥ B → A ⊥ B̄) are a standard exam question. Know the partitioning trick cold.
⚠️ You will be given probability values and asked to determine independence. The workflow is always: find P(A ∩ B), compute P(A) · P(B), compare.
⚠️ De Morgan's law is frequently tested indirectly: they give you P(Eᶜ ∩ Fᶜ) and expect you to convert it to P(E ∪ F) before proceeding.
⚠️ The P(A) = 1 edge case (certain event is independent of everything) appears on exams because it tests whether you can handle the definition mechanically rather than relying on intuition alone.
True or false: If A and B are mutually exclusive and both have nonzero probability, they are independent.
Fill in the blank: P(A ∩ B̄) = P(A) − ___.
True or false: If P(A) = 1, then A is independent of every event B.
Fill in the blank: P(Eᶜ ∩ Fᶜ) = 1 − ___.
True or false: To test independence, it is enough to check whether P(A) = P(B).
Answers: 1. False. 2. P(A ∩ B). 3. True. 4. P(E ∪ F). 5. False.
Q: Given P(A) = 0.5, P(B) = 0.5, and P(A ∪ B) = 0.75, are A and B independent? Show your work.
A: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.5 + 0.5 − 0.75 = 0.25. P(A) · P(B) = 0.25. Since 0.25 = 0.25, A and B are independent.
Q: Prove that if A and B are independent, then A and B̄ are also independent.
A: P(A ∩ B̄) = P(A) − P(A ∩ B) = P(A) − P(A)P(B) = P(A)(1 − P(B)) = P(A)P(B̄). Since the product rule holds for A and B̄, they are independent.
Q: If P(A) = 1, why must A be independent of any event B?
A: P(A ∩ B) ≥ P(B) − P(Ā) = P(B) − 0 = P(B). But P(A ∩ B) ≤ P(B) as well. So P(A ∩ B) = P(B) = 1 · P(B) = P(A) · P(B).
Q: Given P(E) = 0.4, P(F) = 0.5, and P(Eᶜ ∩ Fᶜ) = 0.3, determine whether E and F are independent.
A: By De Morgan, P(E ∪ F) = 1 − P(Eᶜ ∩ Fᶜ) = 1 − 0.3 = 0.7. Then P(E ∩ F) = 0.4 + 0.5 − 0.7 = 0.2. And P(E) · P(F) = 0.2. Equal, so E and F are independent.
Q: You know P(I|J) = 0.7 and P(Iᶜ|Jᶜ) = 0.3. What can you deduce about the independence of I and J?
A: P(Iᶜ|Jᶜ) = 0.3, so P(I|Jᶜ) = 1 − 0.3 = 0.7. Since P(I|J) = P(I|Jᶜ) = 0.7, knowing J does not change the probability of I. Therefore I and J are independent. (You can also verify: P(I) = 0.7 by the law of total probability, and P(I ∩ J) = 0.7 · P(J) = P(I) · P(J) for any value of P(J).)
This connects to conditional probability and Bayes' theorem because independence simplifies the conditional probability formula: if A ⊥ B, then P(A|B) = P(A). Many Bayes' theorem problems begin by checking whether the prior and the evidence are independent or not.
It also connects to combinatorics and counting: when you draw with replacement, successive draws are independent, which lets you multiply probabilities. Drawing without replacement creates dependence, and that changes the calculation entirely.
Later in the course, independence will underpin the definition of random samples and the derivation of sampling distributions.
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