Answer:
Given,\[P\,(A)=\frac{1}{2},\]\[P\,(B)=\frac{1}{3}\]and \[P\,(A\cup B)=\frac{2}{3}\] \[\therefore \] \[P\,(A\cap B)=P\,(A)+P\,(B)-P\,(A\cup B)\] \[=\frac{1}{2}+\frac{1}{3}-\frac{2}{3}=\frac{3+2-4}{6}=\frac{1}{6}\] Here, \[P\,(A)\cdot P\,(B)=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}\] \[\therefore \]\[P\,(A)\cdot P\,(B)=P\,(A\cap B)\] \[\therefore \]Events A and B are mutually independent.
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