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) =