Let S={1,2,3, ldots, 2022}. Then the probabil | vedclass

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Question

Total number of elements $=2022$

$2022=2 	imes 3 	imes 337$

$\operatorname{HCF}( n , 2022)=1$

is feasible when the value of ' $n$ '

Vedclass · Accepted Answer

$\frac{112}{337}$

Vedclass · Answer

Total number of elements $=2022$

$2022=2 	imes 3 	imes 337$

$\operatorname{HCF}( n , 2022)=1$

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 $=\frac{672}{2022}=\frac{112}{337}$

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

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