Probability Question : What is the probability of Attendance of either principal or assistant principal only? 
Three people are working at a school, namely the principal, the assistant principal, and the secretary. If the probability of attendance of the principal equals the probability of attendance of the assistant principal $=0.9$, and the probability of attendance of the secretary $=0.8$, then what is the probability of attendance of either principal or assistant principal only? 

My try:
$$P(P\text{ or }As)=0.9+0.8-(0.9 \cdot 0.8)=0.98$$
Is that okay?
 A: Letting $P$ denote the event that the principal attends is potentially confusing since we also use $P(E)$ to denote the probability of an event $E$. So let's let $E_P$ be the event that the principal attends. Let's let $E_A$ be the event that the assistant principal attends, and let's let $E_S$ be the event that the secretary attends.
In your solution, you used the idea that $P(E_P\text{ and }E_A)=P(E_P)\cdot P(E_A)$. Note that this is only true if $E_P$ and $E_A$ are independent. The original post does not mention whether or not these events are independent. Were we given that $E_P$ and $E_A$ are independent? If $E_P$ and $E_A$ are not independent, then we will need additional information to solve this problem.
From here on, let's suppose that $E_P$, $E_A$, and $E_S$ are independent.
In your solution, you used the fact that
$$P(E_A\text{ or }E_P)=P(E_A)+P(E_P)-P(E_P\text{ and }E_A)$$
to calculate $P(E_A\text{ or }E_P)$. Note that "$E_P\text{ or }E_A$" is the event that either $E_P$ happens, or $E_A$ happens, or $E_P$ and $E_A$ both happen. The question in the original post asks for the "probability of attendance of either principal or assistant principal only?" Hence I think the event we are looking for is
$$(E_P\text{ and not }E_A\text{ and not }E_S)\text{ or }(\text{not }E_P\text{ and }E_A\text{ and not }E_S).$$
Let's let $E_1=(E_P\text{ and not }E_A\text{ and not }E_S)$ and let $E_2=(\text{not }E_P\text{ and }E_A\text{ and not }E_S)$.
Note that
$$\begin{align*}
P(E_1) &= P(E_P\text{ and not }E_A\text{ and not }E_S) \\
&= P(E_P)\cdot P(\text{not }E_A)\cdot P(\text{not }E_S) \\
&= 0.9\cdot0.1\cdot0.2 \\
&= 0.018
\end{align*}$$
Similarly
$$\begin{align*}
P(E_2) &= P(\text{not }E_P\text{ and }E_A\text{ and not }E_S) \\
&= P(\text{not }E_P)\cdot P(E_A)\cdot P(\text{not }E_S) \\
&= 0.1\cdot0.9\cdot0.2 \\
&= 0.018
\end{align*}$$
Finally, note that $E_1$ and $E_2$ are mutually exclusive. Hence
$$P(E_1\text{ or }E_2)=P(E_1)+P(E_2)=0.018+0.018=0.036.$$
A: $P$(Principal only)$=0.9×0.1×0.2=.018$
since Principal and Assistant principal have eqaul attendance probabilities,
$P$(principal only)$=P$(assistant principal only)$=0.018$
Summing up, the required probability is
$P=0.036$
