Q.Events A and B are such that $P(A) = \frac{1}{2}$, $P(B) = \frac{7}{12}$ and P(not A or not B) = $\frac{1}{4}$. State whether A and B are independent ?

Concept: Event Independence — Two events are independent iff $P(A \cap B) = P(A) \cdot P(B)$.

Step 1: Use the given $P(\text{not }A \text{ or not }B) = \frac{1}{4}$. By De Morgan’s law, $\text{not }A \text{ or not }B = \overline{A \cap B}$, so

$P(\overline{A \cap B}) = \frac{1}{4}$.

Step 2: Hence $P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$.

Step 3: Compute $P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{7}{12} = \frac{7}{24}$. …

1. Use the given probability. By De Morgan's law,

Hence

2. Compute $P(A)\cdot P(B)$.

$P(A)\cdot P(B) = \frac12 \times \frac{7}{12} = \frac{7}{24}.$ …

Method: Deciding independence from union/complement data

When a question gives $P(A)$, $P(B)$ and a compound probability (like "not A or not B") and asks whether the events are independent, the technique is: recover $P(A \cap B)$ first, then apply the product test.

Steps

Step 1: Simplify the given compound event to an intersection or its complement.

Translate the words with De Morgan. For example "not A or not B" is

Step 2: Compute the product $P(A)\,P(B)$.

This is what the intersection would equal if the events were independent. …

Common Mistakes

Mistake 1: Misreading "not $A$ or not $B$" as $(A\cup B)'$.

Why it's wrong: De Morgan gives $A'\cup B'=(A\cap B)'$, so it equals $1-P(A\cap B)$ — that is what lets you recover $P(A\cap B)$. Confusing it with $(A\cup B)'$ produces the wrong intersection and a wrong verdict.

Mistake 2: Declaring the events independent (or not) without actually comparing $P(A\cap B)$ with $P(A)P(B)$. …