Given, Sample space is the set of first 1000 natural numbers.

∴ n(S) = 1000 

Let A be the event of choosing the number such that it is multiple of 2 

∴ n(A) = [1000/2] = [500] = 500 {where [.] represents Greatest integer function}

∴P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{500}{1000}\) 

Let B be the event of choosing the number such that it is multiple of 9 

∴ n(B) = [1000/9] = [111.11] = 111 {where [.] represents Greatest integer function}

∴P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{111}{1000}\) 

We need to find the P(such that number chosen is multiple of 2 or 9) 

∵ 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 number is a multiple of both 2 and 9 or we can say that it is a multiple of 18.

n(A∩B) = [1000/18] = [55.55] = 55

∴ P(A ∩ B) = \(\frac{n(A∪B)}{n(s)}\) = \(\frac{55}{1000}\)

∴ P(A ∩ B) =   \(\frac{500}{1000}+\frac{111}{1000}-\frac{55}{1000}\) 

= \(\frac{556}{1000}=\frac{139}{250}​​ ​\)