Question

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

Answer 1

Given, $P\left(notAornotB\right)=\frac{1}{4}\Rightarrow P\left(A^{\prime }or{B}^{\prime }\right)=\frac{1}{4}\Rightarrow P(A^{\prime }\cup B^{\prime })=\frac{1}{4}\Rightarrow P\left[\right(A\cap B\left)\right]=\frac{1}{4}[\because P(A^{\prime }\cup B^{\prime })=P(A\cap B{)}^{\prime }]\Rightarrow 1-P(A\cap B)=\frac{1}{4}\Rightarrow P(A\cap B)=1-\frac{1}{4}=\frac{3}{4}Also,P(A)\times P(B)=\frac{1}{2}\times \frac{7}{12}=\frac{7}{24}\ne \frac{3}{4}[Given,P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{7}{12}]\Rightarrow P(A)\times P(B)\ne P(A\cap B)$
$\therefore$ A and B are not independent events.