Q13:

If A and B are any two events such that $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then

Step 1: Formal Representation of the Given Condition

The problem provides the following equation involving the probabilities of two events, $A$ and $B$, within a sample space $S$:

$P(A) + P(B) - P(A \cap B) = P(A)$

[Given: The probability of the union of two events is defined by the Addition Theorem of Probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$].

Step 2: Algebraic Simplification

By substituting the definition of the union of two events into the given equation, we derive the following logical equivalence:

$P(A \cup B) = P(A)$

[By the Axiom of Probability, for any two sets $A$ and $B$, $A \subseteq (A \cup B)$].

Step 3: Logical Deduction of Set Relations

If $P(A \cup B) = P(A)$, this implies that the event $(A \cup B)$ is effectively equivalent to event $A$ in terms of probability measure. Since $A \subseteq (A \cup B)$ is always true, the condition $P(A \cup B) = P(A)$ implies that the set $B$ must be a subset of $A$.

Specifically:

• If $B \subseteq A$, then $A \cup B = A$.
• Consequently, $P(A \cup B) = P(A)$.

[Per the Monotonicity Property of Probability: If $B \subseteq A$, then $P(B) \leq P(A)$ and $P(A \cap B) = P(B)$].

Step 4: Verification via the Inclusion-Exclusion Principle

Revisiting the original equation: $P(A) + P(B) - P(A \cap B) = P(A)$.

Subtract $P(A)$ from both sides:

$P(B) - P(A \cap B) = 0$

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

This result is the necessary and sufficient condition for $B \subseteq A$. If the probability of the intersection of two events is equal to the probability of one of the events, that event must be contained within the other.

Step 5: Synthesis of Results

The given equation simplifies to $P(B) = P(A \cap B)$, which holds if and only if $B$ is a subset of $A$ ($B \subseteq A$).

Final Solution: B \subseteq A