Let's consider the following events
$$A =$$ Integer chosen is a multiple of $$2$$
So, $$A$$ will contain $$A=\{2,4,6,...,100\}$$
We have, $$P\left(A\right)=\dfrac{50}{100}=\dfrac{1}{2}$$
$$B =$$ Integer chosen is a multiple of $$9$$.
So, $$B$$ will contain $$B=\{9,18,27,...,99\}$$
Hence, $$ P\left(B\right)=\dfrac{11}{100}$$
And $$ P\left(A\cap B\right)$$ Integer chosen is a multiple of both $$2$$ and $$9$$
So, $$\left(A\cap B\right)$$ will contain $$P\left(A\cap B\right)=\{18,36,54,72,90\}$$
Hence, $$P\left(A\cap B\right)=\dfrac{5}{100}$$
We need to calculate the probability of event where the Integer is a multiple of either $$2$$ or $$9$$
So, the required probability $$= P\left(A\cup B\right) = P\left(A\right)+ P\left(B\right)- P\left(A\cap B\right)$$
$$\Rightarrow \dfrac{1}{2}+\dfrac{11}{100}-\dfrac{5}{100}$$
$$\Rightarrow \dfrac{50+11-5}{100}$$
$$\Rightarrow \dfrac{56}{100}$$