Exam PProbability foundationsFree to read

Conditional probability and independence

Conditioning shrinks the sample space to B and renormalises; independence is the special case where that shrinking changes nothing.

The formulas

Definition

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)},P(B)>0$$

Multiplication rule

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

Independence

$$P(A\cap B)=P(A)P(B)⟺P(A\mid B)=P(A)$$

Chain rule

$$P(A_1\cap \cdots \cap A_n)=P(A_1)P(A_2\mid A_1)\cdots P(A_n\mid A_1\cap \cdots \cap A_{n-1})$$

Where it comes from

1. Once $B$ is known to have happened, only outcomes inside $B$ remain possible, so probabilities must be rescaled by $1/P(B)$ to sum to 1 again.
2. The multiplication rule is that definition rearranged, and it is usually the more useful form on the exam.
3. Independence says the conditional and unconditional probabilities agree — knowing $B$ tells you nothing about $A$.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• The thing after the bar is the new denominator.
• Independent means multiply; disjoint means add. They are opposite instructions and both are cued by 'and'/'or'.

Traps

• Reading a joint probability as a conditional one.
• Assuming P(A|B) = P(B|A). They differ by the ratio of the marginals — that is exactly Bayes.

Related

• Bayes' theorem and the law of total probability
• Probability axioms and set identities
• Joint, marginal and conditional distributions

Drill this: the Exam P question bank has original questions on this topic, and today’s free round is open to everyone.