Question

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

Answer 1

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) = \([\frac{200}{6}] \)= [33.334] = 33 {where [.] represents Greatest integer function} 

∴ P(A) = \(\frac{n(A)}{n(S)} =\frac{33}{200}\)

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

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

∴ P(B) = \(\frac{n(B)}{n(S)} = \frac{25}{200}\) 

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) = \([\frac{200}{24}]\) = [8.33] = 8 

∴ P(A∩B) = \(\frac{nP(A ∩ B)}{n(S)} =\frac{8}{200}\) 

∴ P(A∪B) = \(\frac{33}{200}+\frac{25}{200}-\frac{8}{200}\) 

= \(\frac{5}{200}= \frac{1}{4}\)