Conditional Probability

An underwriter gets a positive genetic test. The population mortality rate is 2%, but this applicant is not randomly drawn, the test changed the information set. $P(\text{disease} \mid \text{positive})$ is not $P(\text{positive} \mid \text{disease})$. Confusing the two is the prosecutor's fallacy.

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

Restrict the sample space to outcomes where $B$ occurred. Then measure how much of that restricted space is also in $A$.

HIGH-FREQUENCY: The exam tests this both ways, computing $P(A \mid B)$ from joint and marginal, and computing $P(A \cap B)$ from a conditional and marginal.

Rearranging gives the multiplication rule:

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

Conditional probability is a valid probability measure, all three axioms hold:

• $P(A \mid B) \geq 0$
• $P(B \mid B) = 1$
• Countable additivity for mutually exclusive events

Common mistakes

• Confusing $P(A \mid B)$ with $P(B \mid A)$. $P(D \mid M_2) = 0.05$ (defect rate). $P(M_2 \mid D) = 0.5263$ (source probability). Answering 0.05 when asked for the source probability is the prosecutor's fallacy. Trap: 0.05.
• Dividing by the wrong marginal. $P(A \mid B) = P(A \cap B)/P(B)$, not $P(A \cap B)/P(A)$. The denominator is always the conditioning event. Trap: $P(H \cap C)/P(H) = 0.30$ instead of $P(H \cap C)/P(C) = 0.60$.
• Not computing $P(B)$ via total probability. You need the denominator. If you only have conditional probabilities and class sizes, you must sum across the partition first.

Bottom line

• Definition: $P(A \mid B) = P(A \cap B)/P(B)$; the denominator is always the conditioning event.
• Multiplication rule: $P(A \cap B) = P(A \mid B) \cdot P(B)$.
• $P(A \mid B) \neq P(B \mid A)$ in general; equating them is the prosecutor's fallacy.
• Conditional probability is itself a valid probability measure: complement, inclusion-exclusion, and Bayes all apply within it.

Exam shortcut

When a problem gives conditional probabilities and asks for a "reverse" conditional, you need Bayes' theorem. Start by computing the denominator via total probability. Setting up $P(B) = \sum P(B \mid A_i)P(A_i)$ correctly is 80% of the work. "Given = Given the denominator." The event after the bar ($B$) is always the denominator. Prosecutor's fallacy check: Before writing your answer, ask: "Am I computing P(disease | positive) or P(positive | disease)?"

The full lesson (about 3,462 words, 23 min read) adds 5 worked examples, all 8 common mistakes, a self-check, free in the app.

Learning objectives

• 1f

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