A number is chosen from the first 100 natural numbers. The probability of it is a multiple of 4 or 6 is

A is an event that multiple of 4

B is an event that multiple of 6

A = {4, 8, 12, 16, … 100} $\Rightarrow$n(A) = 25

B = {6, 12, 18, … 96} $\Rightarrow$n(B) = 16

$\begin{array}{l}A\cap B=\{12,24,\cdots 96\}\Rightarrow n(A\cap B)=8\\ P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\\ =\frac{25}{100}+\frac{16}{100}-\frac{8}{100}\\ =\frac{33}{100}\end{array}$