Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to:

Total number of 6 digited numbers from0, 1, 2, 3, 4, 5, 6
$= 6\times 6\times 5\times 4\times 3\times 2 n\left(S\right)=6\times 6! 0+1+2+3+4+5+6=21$
Therefore 6 digited number divisible by 3 can be formed by using the digits  1,2,3,4,5,6 in 6! ways  by using the digits 0,1,2,4,5,6 in $5\times 5!$  ways and by using the digits 0,1,2,3,4,5 in $5\times 5!$  ways  $P\left(E\right)=\frac{6!+2\times 5\times 5!}{6\times 6!}$$=\frac{6+10}{36} =\frac{4}{9}$