Question

Let $A$ be a set of all $4-$digit natural numbers whose exactly one digit is $7.$ Then the probability that a randomly chosen element of $A$ leaves remainder $2$ when divided by $5$ is:

• A. $\frac{1}{5}$
• B. $\frac{2}{9}$
• C. $\frac{97}{297}$
• D. $\frac{122}{297}$

Answer 1

The correct option is C $\frac{97}{297}$

Total cases
$={}^4{C}_1\times 9\times 9\times 9-{}^3{C}_1\times 9\times 9$
(as $4$ digit number having $0$ at thousands place have to be excluded)

For a number to have remainder $2$ when divided by $5$ it’s unit digit should be $2$ or $7$

Case $1:$ when unit digit is $2$
Number of four digit number $={}^3{C}_1\times 9\times 9-{}^2{C}_1\times 9$

Case $2:$ when unit digit is $7$
Number of four digit number $=8\times 9\times 9$
So, total number favorable cases $=3\times 9^2-2\times 9+8\times 9^2$
$\therefore$ required Probability $=\frac{(3\times 9\times 9)-(2\times 9)+(8\times 9\times 9)}{(4\times 9^3)-(3\times 9^2)}$
$=\frac{97}{297}$