A number is chosen at random from $1$ to $200$. What is the probability that it is a perfect square or a multiple of $6$?

Total numbers from $1$ to $200$, $n(S)=200$.
Let $A$ be the set of perfect squares from $1$ to $200$:
The perfect squares are $1^2,2^2,3^2,\dots ,14^2$ (since $14^2=196$ and $15^2=225$).
So, $n(A)=14$.

Let $B$ be the set of multiples of $6$ from $1$ to $200$:

$$\text{Number of multiples}=\lfloor \frac{200}{6}\rfloor =33$$

So, $n(B)=33$.

Let $A\cap B$ be the set of numbers that are both perfect squares and multiples of $6$.
For a perfect square to be a multiple of $6$, it must be a multiple of $\text{LCM}(2,3)=6$ in its base, meaning it must be a multiple of $6^2=36$.
Perfect squares that are multiples of $36$ up to $200$ are:
$36\times 1=36$ ($6^2$)
$36\times 4=144$ ($12^2$)
(The next one is $36\times 9=324>200$)
So, $n(A\cap B)=2$.

By the principle of inclusion-exclusion, the total number of favorable outcomes is:

$$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$

$$n(A\cup B)=14+33-2=45$$

The required probability is:

$$\text{Probability}=\frac{n(A\cup B)}{n(S)}=\frac{45}{200}=\frac{9}{40}$$