An integer is chosen at random from first 200 positive integers. Find the probability that the integer is divisible by 6 or 8.

Given, Sample space is the set of first 200 natural numbers.


n(S) = 200


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


n(A) = [200/6] = [33.334] = 33 {where [.] represents Greatest integer function}


P(A) =


Let B be the event of choosing the number such that it is divisible by 8


n(B) = [200/8] = [25] = 25 {where [.] represents Greatest integer function}


P(B) =


We need to find the P(such that number chosen is divisible by 6 or 8)


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 24.


n(A B) = [200/24] = [8.33] = 8


P(A B) =


P(A B) =


23