One number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

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) = [100/4] = [25] = 25 {where [.] represents Greatest integer function}


P(A) =


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


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


P(B) =


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) = [100/12] = [8.33] = 8


P(A B) =


P(A B) =


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