# Events A and B are such that

Question:

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?

Solution:

It is given that $\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{7}{12}$, and $\mathrm{P}($ not $\mathrm{A}$ or not $\mathrm{B})=\frac{1}{4}$

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

$\Rightarrow P\left((A \cap B)^{\prime}\right)=\frac{1}{4}$                                $\left[A^{\prime} \cup B^{\prime}=(A \cap B)^{\prime}\right]$

$\Rightarrow 1-P(A \cap B)=\frac{1}{4}$

$\Rightarrow P(A \cap B)=\frac{3}{4}$                                                                                                                         ...(1)

However, $\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \cdot \frac{7}{12}=\frac{7}{24}$     ...(2)

Here, $\frac{3}{4} \neq \frac{7}{24}$

$\therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B}) \neq \mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$

Therefore, A and B are not independent events.