Given, Sample space is the set of first 100 natural numbers. 

∴ n(S) = 100

Let A be the event of choosing the number such that it is divisible by 4 

∴ n(A) = \([\frac{100}{4}] \) = [25] = 25 {where [.] represents Greatest integer function} 

∴ P(A) =   \(\frac{n(A)}{n(S)}\) = \(\frac{25}{100} =\frac{1}{4}\) 

Let B be the event of choosing the number such that it is divisible by 6 

∴ n(B) = \([\frac{100}{6}]\)  = [16.67] = 16 {where [.] represents Greatest integer function} 

∴ P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{16}{100} =\frac{4}{25}\) 

We need to find the P(such that number chosen is divisible by 4 or 6) 

∵ P(A or B) = P(A∪B) 

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that: 

P(E∪F) = P(E) + P(F) – P(E∩F)

∴ P(A∪B) = P(A) + P(B) – P(A∩B) 

We don’t have value of P(A∩B) which represents event of choosing a number such that it is divisible by both 4 and 6 or we can say that it is divisible by 12. 

n(A∩B) = \([\frac{100}{12}] \) = [8.33] = 8 

∴ P(A∩B) = \(\frac{n(A ∩ B)}{n(S)}\) = \(\frac{8}{100} = \frac{2}{25}\) 

∴ P(A∪B) = \(\frac{1}{4}+\frac{4}{25}-\frac{2}{25}\) = \(\frac{1}{4}+\frac{2}{25}\) =\(\frac{33}{100}\)