Question

A natural number $X$ is chosen at random from the first $120$ natural numbers and it is observed that it is 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 $\frac{2}{3}$

$X$ is chosen at random from $\{1,2,3,\cdots ,120\}$
Let $A:X$ is divisible by $8$
$A=\{8,16,24,\cdots 120\}$
$P\left(A\right)=\frac{15}{120}$
$B:X$ is divisible by $6$
$B=\{6,12,18,\cdots 120\}$
$\overline{B}=\{1,2,3,4\cdots 120\}-\{6,12,18,\cdots 120\}$
$\overline{B}\cap A=\{8,16,32,40,56,64,80,88,104,112\}$
$P(\overline{B}\cap A)=\frac{10}{120}$
$\Rightarrow P(X$ is not divisible by $6$ when it is divisible by $8)$
$=P\left(\frac{\overline{B}}{A}\right)=\frac{P(\overline{B}\cap A)}{P\left(A\right)}=\frac{10/120}{15/120}=\frac{2}{3}$