14.Probability
hard

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:

A

$\frac{128}{1011}$

B

$\frac{166}{1011}$

C

$\frac{127}{337}$

D

$\frac{112}{337}$

(JEE MAIN-2022)

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

Standard 11
Mathematics

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