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

The Correct Option is D

To solve the problem of finding the probability that a randomly chosen number \( n \) from the set \( S = \{1, 2, 3, \ldots, 2022\} \) satisfies \( \text{HCF}(n, 2022) = 1 \), we need to determine the number of integers in the set that are coprime to 2022. This is essentially finding the Euler's Totient Function \(\phi(n)\) for \( n = 2022 \).

First, we find the prime factorization of 2022: 

\( 2022 = 2 \times 3 \times 337 \)

Euler's Totient Function \(\phi(n)\) is given by:

\(\phi(n) = n \left(1 - \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right)\ldots\left(1 - \frac{1}{p_k}\right)\)

where \( p_1, p_2, \ldots, p_k \) are the distinct prime factors of \( n \).

Applying this to \( 2022 = 2 \times 3 \times 337 \), we have:

\(\phi(2022) = 2022 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{337}\right)\)

Calculating each term:

• \(1 - \frac{1}{2} = \frac{1}{2}\)
• \(1 - \frac{1}{3} = \frac{2}{3}\)
• \(1 - \frac{1}{337} = \frac{336}{337}\)

Substituting these into the formula:

\(\phi(2022) = 2022 \times \frac{1}{2} \times \frac{2}{3} \times \frac{336}{337}\)

Simplifying:

\( \frac{2022}{2} = 1011\)

\( \frac{1011 \times 2}{3} = 674\)

\( \frac{674 \times 336}{337} = 672\)

Therefore, there are 672 numbers in the set \( S \) that are coprime to 2022.

The probability is the ratio of coprime numbers to the total numbers in the set:

\(\frac{672}{2022}\)

Simplifying this fraction:

We divide the numerator and the denominator by their greatest common divisor (GCD), which is 6:

\(672 \div 6 = 112\)

\(2022 \div 6 = 337\)

So, the probability is:

\(\frac{112}{337}\)

Thus, the correct answer is: \(\frac{112}{337}\)

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
\(=\frac{2022−1350}{2022}\)
\(=\frac{672}{2022}\)
\(=\frac{112}{337}\)