BITSAT 2025 Shift 1 Paper

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……2022\}$\Rightarrow n(A) \div 2 = 1011$$B=$ Number which are divisible by 3from $\{1,2,3,\ldots,2022\}$$n(B)=674$$A \cap B =$ Number which are divisible by 6from $\{1,2,3,\ldots,2022\}$6,12,18..............,2022Here $n(A \cap B)=337$$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,1022\}$$C=\{337,674,1011,1348,1685,2022\}$counted in counted in counted in Set $(A \cup B) \;\;$ Set $(A \cup B) \;\;$ Set $(A \cup B)$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 $=\;\frac{672}{2022}=\;\frac{112}{337}$