Solution:
Given: $\mathrm{P}(\mathrm{A})=1 / 2, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=1 / 5$ and let the probability of event be $\mathrm{P}(\mathrm{B})=\mathrm{p}$
(i) Mutually exclusive
When A and B are mutually exclusive which implies that if two events cannot occur at the same moment, they are mutually exclusive or discontinuous.
Then $(A \cap B)=\varphi$
$\Rightarrow P(A \cap B)=0$
As we know that, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Substituting the value and evaluating:
$\Rightarrow 3 / 5=1 / 2+p-0$
$\Rightarrow \mathrm{P}=3 / 5-1 / 2=1 / 10$
(ii) Independent
When A and B are independent.Two events are independent, statistically independent, or stochastically independent if the occurrence of one does not affect the probability of occurrence of the other.
$\Rightarrow P(A \cap B)=P(A) \cdot P(B)$
$$
\Rightarrow P(A \cap B)=1 / 2 p
$$
$$
\text { As we know, } P(A \cup B)=P(A)+P(B)-P(A \cap B)
$$
$$
\Rightarrow 3 / 5=1 / 2+2-p / 2
$$
$\Rightarrow \mathrm{p} / 2=3 / 5-1 / 2$
$$
\Rightarrow p=2 \times 1 / 10=1 / 5
$$