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

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) = n (A) / n (S)

= 25/100

= 1/4

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) = n (B) / n (S)

= 16/100

= 4 /25

Now, we need to find the P (such that number chosen is divisible by 4 or 6)

P (A or B) = P (A ∪ B)

By using the 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)

[Since, we don’t have the 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) = n (A ∩ B) / n (S)

= 8/100

= 2/25

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

P (A ∪ B) = ¼ + 4/25 – 2/25

= 1/4 + 2/25

= 33/100