Mathematics - Probability Question with Solution | TestHub

Given,

$S=\left\{1,2,3,\dots ,2022\right\}$

So, total number of elements$=2022$

Now factors of $2022=2\times 3\times 337$

Now $HCF\left(n,2022\right)=1$ is feasible only when the value of '$n$' and $2022$ has no common factor.

Now let $A=$Number which are divisible by $2$ from $\left\{1,2,3,...2022\right\}$

So, $n\left(A\right)=1011$

Now let $B=$Number which are divisible by $3$by $3$ from $\left(1,2,3,...2022\right)$

So, $n\left(B\right)=674$

Now $A\cap B=$Number which are divisible by $6$

Now from $\left\{1,2,3,...2022\right\}$ number multiple of $6$ are $6,12,18,....2022$

So, $n\left(A\cap B\right)=337$

Now applying the formula we get, $n\left(A\cup B\right)=n\left(A\right)+n\left(B\right)-n\left(A\cap B\right)$

$=1011+674-337$

$=1348$

Now let $C=$Number which divisible by $337$ from $\left\{1,.....1022\right\}$

Total elements which are divisible by $2$ or $3$ or $337=1348+2=1350$

Favourable cases$=$elements which are neither divisible by $2,3$ or $337$

$=2022-1350=672$

Required probability$=\frac{672}{2022}=\frac{112}{337}$