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Mathematics · Ch 9 — Probability

Theorem of Total Probability

9.5.2

Theorem of Total Probability

13.5.2 Theorem of Total Probability

Understanding the Need for This Theorem

When we want to find the probability of an event AA, we often know how AA behaves under different conditions or scenarios. For example, the probability that a construction job finishes on time depends on whether there is a strike or not. The Theorem of Total Probability gives us a systematic way to combine these conditional probabilities into one overall probability.

The key idea is to break the sample space into mutually exclusive and exhaustive pieces — a partition — and then express P(A)P(A) as a weighted average of the conditional probabilities P(A∣Ei)P(A|E_i), where the weights are the probabilities P(Ei)P(E_i) of the pieces themselves.

Partition of a Sample Space

Before stating the theorem, recall what a partition means. A collection of events {E1,E2,…,En}\{E_1, E_2, \dots, E_n\} is called a partition of the sample space SS if:

  • Ei∩Ej=ϕE_i \cap E_j = \phi for all i≠ji \neq j (the events are pairwise disjoint — no two overlap)
  • E1∪E2∪⋯∪En=SE_1 \cup E_2 \cup \dots \cup E_n = S (their union covers the entire sample space)
  • P(Ei)>0P(E_i) > 0 for each ii (each event has a non-zero probability of occurring)

In everyday language, a partition divides the sample space into non-overlapping pieces that together account for every possible outcome. Exactly one of the EiE_i must occur.

Statement of the Theorem of Total Probability

Theorem of Total Probability

Let {E1,E2,…,En}\{E_1, E_2, \dots, E_n\} be a partition of the sample space SS, and suppose each EiE_i has non-zero probability. Let AA be any event associated with SS. Then

P(A)=P(E1)P(A∣E1)+P(E2)P(A∣E2)+⋯+P(En)P(A∣En)P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + \dots + P(E_n)P(A|E_n)

or, in compact summation notation,

P(A)=∑j=1nP(Ej)P(A∣Ej)P(A) = \sum_{j=1}^{n} P(E_j) P(A|E_j)

Proof of the Theorem

›Proof

Step 1: Express AA in terms of the partition.

Since S=E1∪E2∪⋯∪EnS = E_1 \cup E_2 \cup \dots \cup E_n, we can write

A=A∩S=A∩(E1∪E2∪⋯∪En)A = A \cap S = A \cap (E_1 \cup E_2 \cup \dots \cup E_n)

Using the distributive law of set operations,

A=(A∩E1)∪(A∩E2)∪⋯∪(A∩En)A = (A \cap E_1) \cup (A \cap E_2) \cup \dots \cup (A \cap E_n)

Step 2: Show these pieces are disjoint.

Because EiE_i and EjE_j are disjoint for i≠ji \neq j, and A∩EiA \cap E_i is a subset of EiE_i while A∩EjA \cap E_j is a subset of EjE_j, it follows that A∩EiA \cap E_i and A∩EjA \cap E_j are also disjoint for all i≠ji \neq j.

Step 3: Apply the addition rule for mutually exclusive events.

Since the events (A∩E1),(A∩E2),…,(A∩En)(A \cap E_1), (A \cap E_2), \dots, (A \cap E_n) are pairwise disjoint,

P(A)=P[(A∩E1)∪(A∩E2)∪⋯∪(A∩En)]P(A) = P[(A \cap E_1) \cup (A \cap E_2) \cup \dots \cup (A \cap E_n)]

P(A)=P(A∩E1)+P(A∩E2)+⋯+P(A∩En)P(A) = P(A \cap E_1) + P(A \cap E_2) + \dots + P(A \cap E_n)

Step 4: Use the multiplication rule of probability.

For each ii, since P(Ei)≠0P(E_i) \neq 0, the multiplication rule gives

P(A∩Ei)=P(Ei)P(A∣Ei)P(A \cap E_i) = P(E_i) P(A|E_i)

Substituting this into the sum,

P(A)=P(E1)P(A∣E1)+P(E2)P(A∣E2)+⋯+P(En)P(A∣En)P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + \dots + P(E_n)P(A|E_n)

which is exactly the statement of the theorem.

The Special Case of Two Events

When the partition consists of just two events EE and E′E' (where E′E' is the complement of EE), the theorem simplifies to:

P(A)=P(E)P(A∣E)+P(E′)P(A∣E′)P(A) = P(E)P(A|E) + P(E')P(A|E')

This two-event form is extremely common in applications — for example, when an outcome depends simply on whether or not a particular event occurs, splitting the sample space into that event and its complement.

Connection to Bayes' Theorem …