If the probability that a number selected from the set {1, 2, 3, …., 1000} is divisible by 3 but neither divisible by 5 nor by 7, is $\frac{m}{n}$ , then (n – m) is (where m and n are relatively prime number)

Let property, divisible by 5 = A n(A) = 200
Divisible by 7 = B  n(B) = 142
Divisible by 3 = C  n(C) = 333
 $n\left(\overline{A}\cap \overline{B}\cap \overline{C}\right)$
=  $n\left(\overline{A}\cup \overline{B}\cap \overline{C}\right)$
 
= $n\left(C\right)-n\left(C\cap A\cup B\right)$
=  $333-\left[n\left(C\cap A\right)+n\left(C\cap B\right)-n\left(C\cap A\cap B\right)\right]$
= 333 – [66 + 47 – 9]
= 333 – 104
= 229
Hence probability =  $\frac{229}{1000}$