Probability · step 2 of 13

Conditional probability & independence

P(A|B) = P(A ∩ B)/P(B) restricts attention to the outcomes where B occurred — it renormalises the sample space to B. Events are independent when P(A ∩ B) = P(A)P(B), equivalently when P(A|B) = P(A): knowing B tells you nothing about A. Independence is a computational claim to be verified, not an assumption to be made from the wording.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Apply the definitionP(A|B) = P(A ∩ B)/P(B), requiring P(B) > 0. In a finite equally likely space this is |A ∩ B|/|B|.
Use the multiplication rule for sequencesP(A ∩ B) = P(A)P(B|A). Chain it for longer sequences — this is what tree diagrams compute.
Test independence numericallyCompare P(A ∩ B) with P(A)P(B). Equality means independent; anything else means dependent.
Distinguish independence from disjointnessDisjoint events with positive probability are never independent, since P(A ∩ B) = 0 ≠ P(A)P(B).

Worked example

A card is drawn from a standard pack. Are "the card is a king" and "the card is a heart" independent?

P(king) = 4/52 = 1/13. P(heart) = 13/52 = 1/4.
P(king and heart) = 1/52, since there is exactly one king of hearts.
Product: (1/13)(1/4) = 1/52.
The two agree.

Answer. Independent — P(K ∩ H) = P(K)P(H) = 1/52. Knowing the suit tells you nothing about the rank.

Where marks get dropped

These are the specific errors that cost credit on conditional probability & independence questions — QED's rubric penalises each of them separately.

Confusing P(A|B) with P(B|A). They differ whenever P(A) ≠ P(B), and conflating them is the base rate fallacy.
Assuming independence because the events "feel unrelated". Verify it with the product rule.
Treating mutually exclusive as independent. They are opposite extremes of dependence.

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Conditional probability & independence — frequently asked questions

What is pairwise versus mutual independence?

Pairwise means every pair is independent; mutual requires every subset to factor. Pairwise does not imply mutual — there are standard three-event counterexamples with coin flips.

Does P(A|B) = P(B|A) ever hold?

Yes, exactly when P(A) = P(B), which follows immediately from Bayes. It is a coincidence of the numbers, not a general rule.

Why must P(B) > 0?

Because conditioning divides by P(B). Conditioning on a probability-zero event needs measure-theoretic machinery beyond a first course.

The rest of Probability

Events, conditional probability, random variables. Each subtopic below has its own method, worked example and mark-losing traps.

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