Q.If A and B are two events such that $P(A) \neq 0$ and $P(B | A) = 1$, then (A) $A \subset B$ (B) $B \subset A$ (C) $B = \phi$ (D) $A = \phi$

Concept: Conditional Probability — $P(B|A) = 1$ means $B$ occurs with certainty given $A$ has occurred.

Step 1:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = 1$

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

Step 2: …

Why This Works: The Intuition Behind Conditional Probability

Conditional probability $P(B|A)$ answers the question: If we know $A$ has happened, what is the chance that $B$ also happens? When that probability is exactly 1, it means that every outcome in $A$ is also an outcome in $B$ — there is no part of $A$ that lies outside $B$.

Think of it this way: if $A$ were a region that extended beyond $B$, then some outcomes in $A$ would not be in $B$, and $P(B|A)$ would be less than 1. The only way to get a perfect 1 is if $A$ is completely contained inside $B$.

This is the definition we will use. Since $P(A) \neq 0$, division is safe.

Step-by-Step Reasoning

1. Write the given condition using the definition. We have $P(B|A) = 1$. By the formula for conditional probability:

2. Multiply both sides by $P(A)$. Since $P(A) \neq 0$, we can multiply through:

3. Interpret this equality.

   $P(A \cap B)$ is the probability that both $A$ and $B$ occur. $P(A)$ is the probability that $A$ occurs. If these two numbers are equal, it means that every outcome that makes $A$ happen also makes $B$ happen — there is no part of $A$ that occurs without $B$.

4. Translate probability into set theory.

   In probability, $P(A \cap B) = P(A)$ implies that the event $A$ is essentially a subset of $B$, except possibly for a set of measure zero. But since we are dealing with events in a standard probability space, this means $A \subseteq B$ (up to a null set, and for all practical purposes in this problem, exactly $A \subseteq B$). …

Method: Reading a conditional-probability value as a set relationship

For multiple-choice questions that state a value of $P(B\mid A)$ (or $P(A\mid B)$) and ask what it implies about the events, convert the value back through the definition instead of guessing.

Steps

Step 1: Write the definition and substitute the given value.

Setting this equal to $1$ (and using $P(A)\neq 0$) gives $P(A \cap B) = P(A)$.

Step 2: Interpret the resulting equation as containment.

$P(A \cap B) = P(A)$ says all of $A$'s probability already lies inside $B$ — every outcome of $A$ is an outcome of $B$ — i.e. $A \subseteq B$. Keep the direction straight: $P(B\mid A)=1$ constrains $A$ (it forces $B$), so $A$ sits inside $B$, not the other way round. …

Common Mistakes

Mistake 1: Concluding $B\subset A$ instead of $A\subset B$.

Why it's wrong: $P(B\mid A)=1$ means $A$ forces $B$ (every outcome of $A$ lies in $B$), so $A$ is inside $B$, not the reverse. Correct approach: from $P(A\cap B)=P(A)$, deduce $A\subseteq B$ — option (A). …