Question

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

• A. $\frac{128}{1011}$
• B. $\frac{166}{1011}$
• C. $\frac{127}{337}$
• D. $\frac{112}{337}$

Answer 1

Answer: (D)

$\frac{112}{337}$

Answer 2

S = {1, 2, 3, …… 2022}
HCF (n, 2022) = 1
⇒ n and 2022 have no common factor
Total elements = 2022
2022 = 2 × 3 × 337
M : numbers divisible by 2.
{2, 4, 6, ….., 2022} n(M) = 1011
N : numbers divisible by 3.
{3, 6, 9, ….., 2022} n(N) = 674
L : numbers divisible by 6.
{6, 12, 18, ….., 2022} n(L) = 337
n(M∪N) = n(M) + n(N) –n(L)
= 1011 + 674 – 337
= 1348
0 = Number divisible by 337 but not in M∪N
{337, 1685}
Number divisible by 2, 3 or 337
= 1348 + 2 = 1350
Required probability
$=\frac{2022−1350}{2022}$
$=\frac{672}{2022}$
$=\frac{112}{337}$