**Given: **E_{1} and E_{2} are two independent events such that P(E_{1}) = 0.3 and P(E_{2}) = 0.4

**To Find:**

**(i) **P(E_{1} ∩ E_{2})

We know that,

when E_{1} and E_{2} are independent ,

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

= 0.3 x 0.4

= 0.12

Therefore, P(E_{1} ∩ E_{2}) = 0.12 when E_{1} and E_{2} are independent.

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

We know that,

**Hence,** P(E_{1} \(\cup\) E_{2}) = P(E_{1}) + P(E_{2}) - P(E_{1} \(\cup\) E_{2})

= 0.3 + 0.4 – (0.3 x 0.4)

= 0.58

**Therefore ,** P(E_{1} \(\cup\) E_{2}) = 0.58 when E_{1 }and E_{2} are Independent.

**(iii)** P (\(\overline E_1\cap \overline E_2\)) = P(\(\overline E_1\)) x P(\(\overline E_2\))

since , P(E_{1}) = 0.3 and P(E_{2}) = 0.4

\(\Rightarrow\) P(\(\overline E_1\)) = 1 - - P(E_{1}) = 0.7 and P(\(\overline E_2\)) = 1 - P(E_{2}) = 0.6

**Since, **E_{1} and E_{2} are two independent events

\(\Rightarrow\) \(\overline E_1\) and \(\overline E_2\) are also independent events.

**Therefore,** P(\(\overline E_1\cap \overline E_2\)) = 0.7 x 0.6 = 0.42

**(iv) **P(\(\overline E_1\) \(\cap\) E_{2}) = P(\(\overline E_1\)) x P(E_{2})

= 0.7 x 0.4

= 0.28

**Therefore ,** P(\(\overline E_1\) \(\cap\) E_{2}) = 0.28