TL;DR

If \( \{B_1, B_2, \ldots\} \) is a partition of the sample space, then \( P(A) = \sum_i P(A|B_i)P(B_i) \). This decomposes an unconditional probability via conditioning.

By Valenke Exam Prep Team·Last updated 2026-06-03

Law of Total Probability

If \( \{B_1, B_2, \ldots\} \) is a partition of the sample space, then \( P(A) = \sum_i P(A|B_i)P(B_i) \). This decomposes an unconditional probability via conditioning.

Why it matters for interviews

The primary technique for solving multi-step probability problems in interviews. Combined with Bayes' theorem, it forms the backbone of probabilistic reasoning.

Definition and Mathematical Foundation

If \( \{B_1, B_2, \ldots\} \) is a partition of the sample space, then \( P(A) = \sum_i P(A|B_i)P(B_i) \). This decomposes an unconditional probability via conditioning.

Application in Quantitative Finance

The primary technique for solving multi-step probability problems in interviews. Combined with Bayes' theorem, it forms the backbone of probabilistic reasoning.

Related Concepts

Bayes' Theorem

Related Terms

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Frequently Asked Questions

How is this used in a typical interview problem?▼

Condition on the outcome of the first step. For example, in a dice problem: condition on the first roll, compute P(event | first roll = k) for each k, then weight by 1/6 and sum.

What is the law of total expectation?▼

The analog for expectations: \( E[X] = E[E[X|Y]] \). This iterated expectation (tower property) is equally powerful for computing expectations by conditioning.

Does the partition need to be finite?▼

No. The law extends to countable partitions (sums) and continuous conditioning variables (integrals). The key requirement is that the events form a valid partition of the sample space.