Twelve balls are distributed among three axes. The probability that the first box contains 3 balls is:

$\frac{110}{9}(\frac{2}{3})^{10}$

Total arrangement = 3¹² [As each ball have '3' choices]

Favorable ways = (Choose '3' balls out of 12 and place in I^st box) × (Remaining balls each has 2 choices)

$={^{12}C}_3×2^9$ ⇒ P(required) = $\frac{{^{12}C}_3×2^9}{3^{12}}=\frac{110×2^{10}}{3^{12}}$