Independence

Unit 4 · Probability, Random Variables & Distributions

What AP Stats asks here

Two events are independent when knowing one does not change the probability of the other. The recurring AP trap: confusing independence with mutual exclusivity. Two events with positive probability are at opposite extremes of dependence — mutually exclusive events are perfectly informative about each other, while independent events tell you nothing.

Independence definition

Product rule

P(A \cap B) = P(A) \cdot P(B)

Conditional form

P(A \mid B) = P(A)

Easy confusion

Mutually exclusive

P(A \cap B) = 0

With

P(A), P(B) > 0

\text{mutually exclusive } \Rightarrow \text{NOT independent}

AP Tip: Sampling without replacement breaks strict independence. The 10% rule of thumb says you can treat sequential draws as approximately independent when the sample is less than 10% of the population.

Caution: Independent events have no link; mutually exclusive events have the strongest possible link (one occurring guarantees the other did not). Do not conflate them.

Type 1

Verify independence via product rule

Compute the product of marginals and compare to the joint. Equality is the independence check.

Example 1

P(A) = 0.4

P(B) = 0.3

P(A \cap B) = 0.12

. Are A and B independent? (A) No — the product rule fails. (B) Yes —

P(A) \cdot P(B) = 0.12 = P(A \cap B)

. (C) Cannot tell without

P(A \cup B)

. (D) No — independent events must have

P(A) = P(B)

Practice more of this type— AI-generated · always-new problems

Generate Problems →

Type 2

Independent vs mutually exclusive

If both events have positive probability, they cannot be both independent and mutually exclusive — the two definitions force opposite values for $P(A \cap B)$ .

Example 2

P(A) = 0.5

P(B) = 0.3

, and we are told the events are mutually exclusive. Can they also be independent? (A) Yes — any disjoint events are independent. (B) Yes — mutually exclusive guarantees independence. (C) No — mutually exclusive ⇒

P(A \cap B) = 0

, but independence ⇒

P(A \cap B) = 0.15

. Both cannot hold with

P(A), P(B) > 0

. (D) Only if

P(A) = P(B)

Practice more of this type— AI-generated · always-new problems

Generate Problems →

Type 3

Chain of independent events

For independent trials, multiply probabilities for joint occurrence, or use the complement for 'at least one'.

Example 3

A factory produces parts; each is defective with probability 0.02 independently. Three parts are selected. What is

P(\text{at least one defective})

? (A) 0.06 (B) ≈ 0.0588 (C) ≈ 0.9412 (D) ≈ 0.000008

Practice more of this type— AI-generated · always-new problems

Generate Problems →