Given two mutually exclusive events $A$ and $B$ such that $P(A)=0.45$, $P(B)=0.35$, then $P(A\text{ or }B)$ is
A. $0.1$
B. $0.25$
C. $0.15$
D. $0.8$

Hint: In this question, the probability of the union of two events is to be determined. Since the given events are mutually exclusive, the sets are disjoint sets i.e., the intersection between those events is null. By using the addition theorem on probability, the required probability is calculated.

Formula used: The probability is calculated by,
\[P(E)=\dfrac{n(E)}{n(S)}\]
\[n(E)\] - favourable outcomes and \[n(S)\] - sample.
If there are two events in a sample space, then the addition theorem on probability is given by
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
If the sets are mutually exclusive i.e., $P(A\cap B)=\Phi $ then,
$P(A\cup B)=P(A)+P(B)$

Complete step by step solution: Consider two events $A$ and \[B\] which are mutually exclusive.
Given that,
$P(A)=0.45$
\[P(B)=0.35\]
Since the given events are mutually exclusive,
 i.e., empty or zero.
According to the addition theorem on probability,
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
Then,
\[\begin{align}
  & P(A\cup B)=P(A)+P(B)-0 \\
 & \Rightarrow P(A\cup B)=P(A)+P(B) \\
\end{align}\]
On substituting, we get
\[\begin{align}
  & P(A\cup B)=P(A)+P(B) \\
 & \text{ }=0.45+0.35 \\
 & \text{ }=0.8 \\
\end{align}\]

Thus, Option (D) is correct.

Note: Here we may go wrong with the calculation of $P(A\cap B)$. Mutually exclusive events are disjoint sets. So, there is no intersection between them. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability.