A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?
A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?

Sample space is the set of first 500 natural numbers.

n (S) = 500

Let β€˜A’ be the event of choosing the number such that it is divisible by 3

n (A) = [500/3]

= [166.67]

= 166 {where [.] represents Greatest integer function}

P (A) = n (A) / n (S)

= 166/500

= 83/250

Let β€˜B’ be the event of choosing the number such that it is divisible by 5

n (B) = [500/5]

= [100]

= 100 {where [.] represents Greatest integer function}

P (B) = n (B) / n (S)

= 100/500

= 1/5

Now, we need to find the P (such that number chosen is divisible by 3 or 5)

P (A or B) = P (A βˆͺ B)

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

P (E βˆͺ F) = P (E) + P (F) – P (E ∩ F)

∴ P (A βˆͺ B) = P (A) + P (B) – P (A ∩ B)

[Since, we don’t have value of P(A ∩ B) which represents event of choosing a number such that it is divisible by both 3 and 5 or we can say that it is divisible by 15.]

n(A ∩ B) = [500/15]

= [33.34]

= 33

P (A ∩ B) = n(A ∩ B) / n (S)

= 33/500

∴ P (A βˆͺ B) = P (A) + P (B) – P (A ∩ B)

= 83/250 + 1/5 – 33/500

= [166 + 100 – 33]/500

= 233/500