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माना $S =\{1,2,3, \ldots, 2022\}$ है। तब समुच्चय $S$ से यादृच्छया चुनी गई एक संख्या $n$ के लिए $HCF$ $( n , 2022)=1$ होने की प्रायिकता है:
$\frac{128}{1011}$
$\frac{166}{1011}$
$\frac{127}{337}$
$\frac{112}{337}$
Solution
Total number of elements $=2022$
$2022=2 \times 3 \times 337$
$\operatorname{HCF}( n , 2022)=1$
is feasible when the value of ' $n$ ' and $2022$ has no common factor.
$A=$ Number which are divisible by $2$ from
$\{1,2,3 \ldots . .2022\}$
$n ( A )=1011$
$B =$ Number which are divisible by 3 by 3
from $\{1,2,3 \ldots . .2022\}$
$n ( B )=674$
$A \cap B=$ Number which are divisible by 6
from $\{1,2,3 \ldots \ldots . .2022\}$
$6,12,18 \ldots \ldots \ldots, 2022$
$337=n(A \cap B)$
$n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$=1011+674-337$
$=1348$
$C =$ Number which divisible by $337$ from
$\{1, \ldots \ldots . .1022\}$
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 $=\frac{672}{2022}=\frac{112}{337}$
