Properties of Conditional Probability

Understanding Conditional Probability: The Core Idea

Before we dive into the properties, recall what conditional probability means. When we write $P(E|F)$, we are asking: If we already know that event F has happened, what is the probability that event E also happens? The sample space effectively shrinks from the original S to just the outcomes in F. The formula that captures this is:

This definition is the foundation for everything that follows. The properties we are about to study show that conditional probability behaves just like ordinary probability — but with the condition acting as a new "universe" of outcomes.

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Property 1: The Certainty of the Sample Space and the Conditioning Event

Statement: For any event F in a sample space S (with $P(F) \neq 0$),

Why this makes sense: If we already know that F has occurred, then the original sample space S is definitely true (since S contains all possible outcomes). Similarly, if F has occurred, then F itself is certainly true. Both probabilities should be 1.

Proof:

For $P(S|F)$:

Here, $S \cap F = F$ because F is a subset of S — every outcome in F is also in S.

For $P(F|F)$:

Since $F \cap F = F$, the numerator equals the denominator.

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Property 2: The Addition Rule for Conditional Probability

Statement: For any events A, B, and F (with $P(F) \neq 0$),

Special case: If A and B are disjoint events (i.e., $A \cap B = \emptyset$), then:

Proof:

Start with the definition of conditional probability for the union:

Now, use the distributive law of set operations: $(A \cup B) \cap F = (A \cap F) \cup (B \cap F)$. This gives:

Apply the ordinary addition rule for probability to the numerator:

But $(A \cap F) \cap (B \cap F) = A \cap B \cap F$. So:

Separate the fraction:

Each term is a conditional probability:

For disjoint A and B: If $A \cap B = \emptyset$, then $A \cap B \cap F = \emptyset$, so $P((A \cap B)|F) = 0$. The formula simplifies to:

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Property 3: The Complement Rule for Conditional Probability

Statement: For any event E and conditioning event F (with $P(F) \neq 0$),

where $E'$ denotes the complement of E (i.e., "not E").

Proof:

We know from Property 1 that $P(S|F) = 1$. Since $S = E \cup E'$ (every outcome is either in E or not in E), we can write:

Now, E and E' are disjoint events (they have no common outcomes). Using the special case of Property 2:

Therefore:

Rearranging gives:

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Summary of Key Ideas

1. Conditional probability $P(E|F)$ redefines the sample space to F. All properties of ordinary probability hold within this reduced space. …