Question

A natural number $x$ is chosen at random from the first $120$ natural numbers and it is observed to be divisible by $8,$ then the probability that it is not divisible by $6$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Numbers which are divisible by both $6$ and $8$ are $24,48,72,96,120$

So, $P$ (Divisible by $8$ not by $6$) $=1-\frac{Prob\left(divisiblebyboth\right)}{Prob\left(divisibleby8\right)}$

Numbers divisible by $8$ are $8,16\dots ,120$

So, $120=8+(n-1)8n=15$

So, $P$ (Divisible by $8$ no by $6$)$=1-\frac{\frac{5}{120}}{\frac{15}{120}}=1-\frac{1}{3}=\frac{2}{3}$