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

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$\frac{143}{1000}$

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(B) The set of numbers is $S = \{1, 2, 3, \ldots, 1000\}$, so the total number of outcomes $n(S) = 1000$.We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.These numbers are of the form $n = 7k + 1$, where $k$ is an integer.For $1 \le n \le 1000$, we have $1 \le 7k + 1 \le 1000$.Subtracting $1$ from all parts: $0 \le 7k \le 999$.Dividing by $7$: $0 \le k \le \frac{999}{7} \approx 142.71$.Since $k$ must be an integer, $k$ can take values from $0, 1, 2, \ldots, 142$.The number of such values is $142 - 0 + 1 = 143$.Thus, the number of favorable outcomes $n(E) = 143$.The probability $P(E) = \frac{n(E)}{n(S)} = \frac{143}{1000}$.

$A$ 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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