Get Answers to all your Questions

header-bg qa

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

Option: 1

\frac{128}{1011}


Option: 2

\frac{166}{1011}


Option: 3

\frac{127}{337}


Option: 4

\frac{112}{337}


Answers (1)

best_answer

Total number of elements =2022

2022=2 \times 3 \times 337

\mathrm{HCF \left(n_{1} 2022\right)=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}

\mathrm{n(A)=1011 }

B=Number which are divisible by 3 from {1,2,3,.........,2022}

\mathrm{n(B)=674 }

\mathrm{A \cap B=}\text{ Number which are tivisible by 6}

\begin{aligned} & \text { from }\{1,2,3, \ldots 2022\} \\ & 6,12,18, \ldots .2022 \\ & \mathrm{337=n(A \cap B) }\\ & \mathrm{n(A \cup B)=n(A)+n(B)-n(A \cap B)} \\ & =1011+674-337 \\ & =1348 \end{aligned}

C=Number which divisible by 337 from {1,..............,1022}

Total elements which are divisible by 2 or 3 or 337\

=1348+2=1350
Favourable cases = Element which are neither dinisible
by 2,3 or 337

=2022-1350=672

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

Posted by

Suraj Bhandari

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE