The probability that when 12 distinct balls are distributed among three distinct boxes, the first will contain exactly three balls is

3 balls for Ist box can be selected in ${ }^{12}{C}_3$ ways, remaining 9 balls can be placed in remaining 2 boxes in $2^9$ ways.
Number of favourable cases ${=}^{12}{C}_9\times 2^9$
Total number of cases $=3^{12}$
Required probability $=\frac{{ }^{12}{C}_9\times 2^9}{3^{12}}$