Q.If $P(A) = \frac{1}{2}$, $P(B) = 0$, then $P(A|B)$ is (A) 0 (B) $\frac{1}{2}$ (C) not defined (D) 1

Concept: Conditional Probability — $P(A|B)$ is defined only when $P(B) > 0$.

Step 1: The formula for conditional probability is

Step 2: Here $P(B) = 0$. Division by zero is undefined in probability theory. …

Why This Problem Tests a Definition, Not a Calculation

Many students see $P(A|B) = \frac{P(A \cap B)}{P(B)}$ and try to plug numbers in. But the denominator here is zero — and division by zero is not allowed in mathematics. The question isn't really about $A$ or $B$; it's about whether the definition of conditional probability applies at all.

Conditional probability $P(A|B)$ answers: "Given that $B$ has happened, what is the chance that $A$ also happens?" If $B$ has zero probability, it can never occur — so the question "given $B$" becomes meaningless. You cannot condition on an impossible event.

Step-by-Step Reasoning

1. Recall the definition of conditional probability. For any two events $A$ and $B$, the conditional probability of $A$ given $B$ is

This formula is valid only when $P(B) > 0$. If $P(B) = 0$, the expression is undefined — you cannot divide by zero.

2. Check the given values.

   We have $P(B) = 0$. That means $B$ is an event that occurs with zero probability — an impossible event (or a null event in the probability space).

3. Apply the definition. …

Method: Checking When a Conditional Probability Is Defined

Use this whenever a question asks for $P(A\mid B)$ and you must decide whether the value exists at all, rather than blindly plugging numbers in.

Steps

Step 1: Write the definition and note its domain

Conditional probability is defined only for a conditioning event of positive probability:

Step 2: Inspect the conditioning event first …

Common Mistakes

Mistake 1: Blindly plugging into the formula and dividing by zero.

Seeing $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$, a student substitutes $P(B)=0$ and reports the answer as $0$. Why it's wrong: division by zero is undefined, not zero. Correct approach: the formula is valid only for $P(B)>0$, so the value is not defined.

Mistake 2: Assuming $P(A\cap B)=0$ forces $P(A\mid B)=0$. …