The number of ways in which 9 distinct balls can be distributed among 4 boxes is 49
The number of ways in which B3 contains 3 balls is C39⋅36
Using the classical definition of probability we can say the required probability =49C39⋅36
=27C39⋅(43)9
=928⋅(43)9⇒k=928≈3.11
Now, consider ∣x−3∣<1
⇒−1<x−3<1
⇒−1+3<x<1+3
⇒2<x<4
∴k satisfies ∣x−3∣<1