**Given:** E_{1} and E_{2 }are two events such that P(E_{1}) = \(\frac{1}{3}\) and P(E_{2}) = \(\frac{3}{5}\)

**To Find: **

**(i) **P(E_{1} ∪ E_{2}) when E_{1} and E_{2} are mutually exclusive.

We know that,

When two events are mutually exclusive P(E_{1 }∩ E_{2}) = 0

**Hence,** P(E_{1 }∪ E_{2}) = P(E_{1}) + P(E_{2})

= \(\frac{1}{3}+\frac{3}{5}\)

= \(\frac{14}{15}\)

**Therefore , **P(E_{1} ∪ E_{2}) = \(\frac{14}{15}\) when E_{1} and E_{2} are mutually exclusive.

**(ii) **P(E_{1} ∩ E_{2}) when E_{1} and E_{2} are independent.

We know that when E_{1} and E_{2} are independent ,

P(E_{1} ∩ E_{2}) = P(E_{1}) x P(E_{2})

= \(\frac{1}{3}\times \frac{3}{5}\)

= \(\frac{1}{5}\)

**Therefore, **P(E_{1} ∩ E_{2}) = \(\frac{1}{5}\) when E_{1 }and E_{2} are independent.