Q80 / 160 · AP EAMCET 2006 · Mathematics

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 :

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

Solution

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 $=\frac{n(E)}{n(S)}$ $=\frac{142}{1000}=\frac{71}{500}$

Hence, the correct answer is option: 'A'.

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