Let S = {1, 2, 3, ..., 2022}. Then the probability, that a randomly chosen number n from the set S such that HCF (n, 2022) = 1, is :
Text Solution
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Total number of elements = 2022
2022 = 2x3x337
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}
n(a) =1011
B = Number which are divisible by 3 by 3 from (1,2,3......2022}
n(b)=674
A 
B = Number which are divisible by 6 from {1,2,3........2022}
6,12,18........., 2022
337 = n(A 
B)
n(A 
B) = n(a) + n(b) - n(A 
B)
= 1011+674-337
=1348
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
divisible by 2, 3 or 337
= 2022- 1350
= 672
Required probability 
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