If A and B are two events such that P(A) ≠ 0 and P(B | A) = 1 then

The correct answer is Option (1) → $A⊂B$

Given:

$P(B|A) = 1$ and $P(A) \ne 0$

From conditional probability:

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

So,

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

This implies that all outcomes of $A$ are contained in $B$.

Therefore,

$A \subseteq B$