Q.If A and B are any two events such that $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then (A) $P(B|A) = 1$ (B) $P(A|B) = 1$ (C) $P(B|A) = 0$ (D) $P(A|B) = 0$

Idea: Recognise the left side as the addition rule, then read off what the condition forces.

The addition rule gives $P(A)+P(B)-P(A\cap B)=P(A\cup B)$, so the equation says

Then …

Reading the condition

We are given

The left-hand side is exactly the addition rule for $P(A\cup B)$, so the equation states $P(A\cup B)=P(A)$. Cancelling $P(A)$ from both sides:

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

The overlap of $A$ and $B$ has the same probability as $B$ itself. Since $A\cap B\subseteq B$ always, equal probabilities mean $B$ contributes nothing outside $A$ — every time $B$ happens, $A$ happens too. In set language, $B\subseteq A$ (up to zero-probability outcomes).

Evaluating the options

Using $P(A\cap B)=P(B)$ (and assuming the events have non-zero probability, as usual in these problems):

• (B) $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(B)}{P(B)}=1$. ✓ …

Method: Spotting a hidden standard identity before computing

Many one-mark questions disguise a familiar rule inside an equation. The technique is to recognise the rule first, simplify, then translate the result into whatever is asked.

Steps

Step 1: Match the expression to a known formula.

Scan the left-hand side for a standard pattern. The combination $P(A) + P(B) - P(A\cap B)$ is exactly the addition theorem

Step 2: Substitute and simplify to a set relation. …

Common Mistakes

Mistake 1: Not recognising $P(A)+P(B)-P(A\cap B)$ as $P(A\cup B)$.

Why it's wrong: without spotting the addition theorem, students cannot simplify and start guessing. Correct approach: rewrite the left side as $P(A\cup B)$, so the equation becomes $P(A\cup B)=P(A)$, forcing $P(A\cap B)=P(B)$.

Mistake 2: Confusing which conditional equals $1$. …