(B) The set of numbers is $S = \{1, 2, 3, \ldots, 1000\}$,so $n(S) = 1000$.We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.These number

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(B) The set of numbers is $S = \{1, 2, 3, \ldots, 1000\}$,so $n(S) = 1000$.We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.These numbers are of the form $n = 7k + 1$ for $k \ge 0$.For $1 \le 7k + 1 \le 1000$,we have $0 \le 7k \le 999$,which implies $0 \le k \le \frac{999}{7} \approx 142.71$.Since $k$ must be an integer,$k \in \{0, 1, 2, \ldots, 142\}$.The number of such values is $142 - 0 + 1 = 143$.The probability is $P = \frac{143}{1000}$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

$A$ number $n$ is chosen at random from $\{1, 2, 3, 4, \ldots, 1000\}$. The probability that $n$ is a number that leaves remainder $1$ when divided by $7$ is:

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A
$\frac{71}{500}$
B
$\frac{143}{1000}$
C
$\frac{72}{500}$
D
$\frac{71}{1000}$

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$A$ number $n$ is chosen at random from $\{1, 2, 3, 4, \ldots, 1000\}$. The probability that $n$ is a number that leaves remainder $1$ when divided by $7$ is:

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