Events $A$ and $B$ are such that $P(A)=1 / 2, P(B)=7 / 12$ and $P($ not $A$ or $n o t B)=1 / 4$. State whether $A$ and $B$ are independent?

Solution:

Given: $P(A)=1 / 2, P(B)=7 / 12$ and $P($ not $A$ or $\operatorname{not} B)=1 / 4$
Concept: Two events are independent, statistically independent, or stochastically independent if the occurrence of one does not affect the probability of occurrence of the other.

Calculation:
Evaluating the value of required parameter

$\Rightarrow P\left(A^{\prime} \cup B^{\prime}\right)=1 / 4$

$\Rightarrow P(A \cap B)^{\prime}=1 / 4$

$\Rightarrow 1-P(A \cap B)=1 / 4$

$\Rightarrow P(A \cap B)=1-1 / 4$

$\Rightarrow P(A \cap B)=3 / 4 \ldots \ldots .(1) .$

And $P$ (A). $P(B)=1 / 2 \times 7 / 12=7 / 24 \ldots . .$ (2)

From (1) and (2) $P(A \cap B) \neq P$ (A). $P(B)$

Final Answer: Therefore, A and B are not independent events.