Question

Let $S=\{1,2,3,\dots ,2022\}$. Then the probability, that a randomly chosen number $n$ from the set S such that $\mathrm{HCF}(n,2022)=1$, is

• A. $\frac{128}{1011}$
• B. $\frac{166}{1011}$
• C. $\frac{127}{337}$
• D. $\frac{112}{337}$

Answer 1

Answer: (D)

Total number of elements $=2022$
$2022=2\times 3\times 337$
$\mathrm{HCF}(n,2022)=1$
is feasible when the value of ' $n$ ' and 2022 has no common factor.
$A=$ Number which are divisible by 2 from $\{1,2,3\dots ..2022\}$
$n(A)=1011$
$B=$ Number which are divisible by 3 by 3
from $\{1,2,3\dots \dots 2022\}$
$n(B)=674$
$A\cap B=$ Number which are divisible by 6
from $\{1,2,3\dots \dots ..2022\}$
$6,12,18\dots \dots ..,2022$
$337=n(A\cap B)$
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$=1011+674-337$
$=1348$
$C=$ Number which divisible by 337 from $\{1,\dots \dots ..1022\}$

Total elements which are divisible by 2 or 3 or 337
$=1348+2=1350$
Favourable cases $=$ Element which are neither divisible by 2,3 or 337
$=2022-1350$
$=672$
Required probability $=\frac{672}{2022}=\frac{112}{337}$