Question

If a number $n$ is chosen at random from the set $\left\{1,2,3,........,1000\right\}$. Then, the probability that $n$ is a number that leaves remainder $1$,when divided by $7$, is

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

Answer 1

The correct option is A

$\frac{71}{500}$

Finding the probability:

Given, total outcome is $\left\{1,2,3,........,1000\right\}=1000$

Favourable outcome, Leaves remainder $1$ when divided by $7$ is $\left\{8,15,22,........,995\right\}$

Let, $a=8,a_n=995\&d=7$

Number of term is $a_n=a+\left(n-1\right)d$

$\Rightarrow$ $995=8+\left(n-1\right)7$

$\Rightarrow$ $n=\frac{987}{7}+1$

$\Rightarrow$ $n=142$

Therefore, favourable outcome is $142$

Required probability is $\frac{\mathbf{fav}\mathbf{our}\mathbf{a}\mathbf{b}\mathbf{l}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}}{\mathbf{to}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{}\mathbf{o}\mathbf{u}\mathbf{t}\mathbf{c}\mathbf{o}\mathbf{m}\mathbf{e}}=\frac{142}{1000}=\frac{71}{500}$

Hence, correct answer is option (A).