Let 9 distinct balls be distributed among 4 boxes, B _ 1 , B _ 2 ,...

The number of ways in which $9$ distinct balls can be distributed among $4$ boxes is ${4}^{9}$

The number of ways in which ${B}_{3}$ contains $3$ balls is $C3 9 \cdot {3}^{6}$

Using the classical definition of probability we can say the required probability $= \dfrac{C3 9 \cdot {3}^{6}}{{4}^{9}}$

$= \dfrac{C3 9}{27} \cdot {\left(\dfrac{3}{4}\right)}^{9}$

$= \dfrac{28}{9} \cdot {\left(\dfrac{3}{4}\right)}^{9} ⇒ k = \dfrac{28}{9} \approx 3 . 11$

Now, consider $| x - 3 | < 1$

$⇒ - 1 < x - 3 < 1$

$⇒ - 1 + 3 < x < 1 + 3$

$⇒ 2 < x < 4$

$∴ k$ satisfies $| x - 3 | < 1$