Answer with explanation:
Consider an example of tossing a dice once
Total favorable outcome =6={1,2,3,4,5,6}
[tex]X_{i}=\text{multiple of 2}={2,4,6}\\\\X_{j}=\text{multiple of 3}={3,6}\\\\P(\text{An Event})=\frac{\text{Total favorable outcome}}{\text{Total favorable outcome}}\\\\P(X_{i}})=\frac{3}{6}\\\\=\frac{1}{2}\\\\P(X_{j}})=\frac{2}{6}\\\\=\frac{1}{3}\\\\\rightarrow X_{i} \cap X_{j}={6}\\\\\rightarrow P(X_{i} \cap X_{j})=\frac{1}{6}\\\\\rightarrow P(X_{i} \cap X_{j})=P(X_{i})\times P(X_{j})[/tex]
Hence proved.
