Q.Prove that

Key Idea: Partition of the Sample Space

The events $B$ and $B'$ are mutually exclusive and exhaustive — they form a partition of the sample space. Any event $A$ can be split into the part inside $B$ and the part outside $B$.

Proof of (i):

Since $A = (A \cap B) \cup (A \cap B')$ and the two intersections are disjoint, by the addition rule of probability:

Proof of (ii):

The union $A \cup B$ can be written as the disjoint union of three parts:

$A \cup B = (A \cap B) \cup (A \cap B') \cup (A' \cap B)$ …

The Core Idea: Partitioning the Sample Space

Probability is about measuring the "weight" of an event within a sample space $S$. A powerful trick is to split $S$ into disjoint pieces whose probabilities add up cleanly. Here, the natural split is by event $B$ and its complement $B'$ — because every outcome either belongs to $B$ or not. This is called a partition of $S$.

Why does this work? Because $A$ is the union of two mutually exclusive parts: the part of $A$ that lies inside $B$, and the part of $A$ that lies outside $B$. Since these two parts cannot overlap (an outcome cannot be both in $B$ and not in $B$), their probabilities simply add.

Proving (i) $P(A) = P(A \cap B) + P(A \cap B')$

1. Visualise the sets. Draw a Venn diagram with two overlapping circles $A$ and $B$. The region $A$ is split into two non-overlapping subregions: the lens where $A$ and $B$ overlap ($A \cap B$), and the crescent of $A$ outside $B$ ($A \cap B'$). These two pieces together cover every outcome in $A$ and share no outcomes.

2. Write the union. Since $A \cap B$ and $A \cap B'$ are disjoint:

3. Apply the probability axiom. For mutually exclusive events, the probability of the union is the sum of the probabilities:

That's it — the identity follows directly from the definition of a partition and the additivity axiom of probability.

Proving (ii) $P(A \cup B) = P(A \cap B) + P(A \cap B') + P(A' \cap B)$

1. Think of the union as three disjoint pieces. The union $A \cup B$ consists of:

   • Outcomes in both $A$ and $B$: $A \cap B$
   • Outcomes in only $A$ (not in $B$): $A \cap B'$
   • Outcomes in only $B$ (not in $A$): $A' \cap B$

   These three regions are mutually exclusive — no outcome can be in more than one of them. Together they cover every outcome that is in $A$ or $B$ (or both).

2. Write the union as a disjoint union:

   $A \cup B = (A \cap B) \cup (A \cap B') \cup (A' \cap B)$ …

Method: Proving Probability Identities by Partitioning into Disjoint Pieces

Use this to prove set-based probability identities — the strategy is always to write an event as a union of disjoint pieces, then add.

Steps

Step 1: Split by an event and its complement.

$B$ and $B'$ partition the sample space, so any event $A$ splits as the disjoint union

Step 2: Use additivity for disjoint events.

For mutually exclusive events probabilities add, giving $P(A)=P(A\cap B)+P(A\cap B')$. …

Common Mistakes

Mistake 1: Writing $P(A\cap B')=P(A)-P(B)$.

Why it's wrong: the part of $A$ outside $B$ is $P(A)-P(A\cap B)$, not $P(A)-P(B)$. Correct approach: subtract the overlap $P(A\cap B)$, which is exactly what identity (i) rearranges to.

Mistake 2: Adding probabilities of pieces that are not disjoint.

Why it's wrong: additivity requires mutually exclusive events; $A\cap B$, $A\cap B'$, $A'\cap B$ must first be shown pairwise disjoint. Correct approach: verify disjointness, then add. …