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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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The correct answer is:
$\frac{71}{500}$
Multiples of 7 in $\{1,2, \ldots, 1000\}$ are $7,14,21, \ldots, 994$.
Let the number of terms be $N$
$\therefore \quad 994=7+(N-1) \cdot 7$
$\Rightarrow \quad \frac{987}{7}=(N-1)$
$\Rightarrow \quad N-1=141$
$\Rightarrow \quad N=142$
$\therefore$ Number of terms which leaves remainder 1 when divided by $7=142$ and $n(S)=1000$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}$
$=\frac{142}{1000}=\frac{71}{500}$
Let the number of terms be $N$
$\therefore \quad 994=7+(N-1) \cdot 7$
$\Rightarrow \quad \frac{987}{7}=(N-1)$
$\Rightarrow \quad N-1=141$
$\Rightarrow \quad N=142$
$\therefore$ Number of terms which leaves remainder 1 when divided by $7=142$ and $n(S)=1000$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}$
$=\frac{142}{1000}=\frac{71}{500}$
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