Q77 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
A number $n$ is chosen at random from the natural numbers 2 to 1001 . The probability that $n$ is a number that leaves remainder 1 when divided by 7 , is
- A$\frac{73}{500}$
- B$\frac{71}{1000}$
- C$\frac{143}{1000}$
- Correct answerD$\frac{71}{500}$
Solution
Total number of numbers from 2 to $1001=1000$
Now, the numbers, which leaves remainder 1 when divided by 7 , are $8,15,22, \ldots, 995$.
Let these are $m$ in counting, then we have
$$
\begin{aligned}
\quad 995=8+(m-1) 7 & \Rightarrow 987=(m-1) 7 \\
\Rightarrow \quad m-1=141 & \Rightarrow m=142 \\
\therefore & \text { Required probability }=\frac{142}{1000}=\frac{71}{500}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.