Mathematics - Probability Question with Solution | TestHub

$$\begin{array}{rl}& P(\overline{A\cup B})=\frac{1}{6},P(A\cap B)=\frac{1}{4}\text{ and }P(\bar{A})=\frac{1}{4}\\ & \Rightarrow P(A\cup B)=5/6P(A)=3/4\\ & \text{ Also }P(A\cup B)=P(A)+P(B)-P(A\cap B)\\ & \Rightarrow P(B)=5/6-3/4+1/4=1/3\\ & P(A)P(B)=3/4-1/3=1/4=P(A\cap B)\end{array}$$ Hence $A$ and $B$ are independent but not equally likely.