Question

A number is chosen at random among the first $120$ natural numbers. The probability of the number chosen being a multiple of $5$ or $15$ is :

• A. $\frac{1}{8}$
• B. $\frac{1}{5}$
• C. $\frac{1}{24}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

The number having unit digit 0 or 5 will be divisible by $5 $
All the numbers divisible by 15 are also divisible by $5 $
In first 120 natural numbers total number of multiple of $5, n(A)=24$ and total number of multiple of $15, n(B)=8$ and $n(A \cap B)=8$
$ \therefore \, n(A \cup B) =n(A)+n(B)-n(A \cap B) $
$=24+8-8=24 $
$\therefore $ Required probability $=\frac{24}{120}=\frac{1}{5}$