If three dices are thrown then the probability that the sum of the numbers on their uppermost faces to be atleast 5 is

If there dice are drawn, then $n\left(\le \right)=6^3=216$
Let E the sum of the number on their uppermost faces are less than 5 as follows
$\left(\mathrm{1,1},1\right),\left(\mathrm{1,1},2\right),\left(\mathrm{1,2},1\right),\left(\mathrm{2,1},1\right)$
$\therefore n\left(E\right)=4$
Now, probability that the sum of the numbers on their uppermost faces to be atleast 5
$=1-P\left(E_2\right)=1-\frac{4}{216}=1-\frac{1}{54}=\frac{53}{54}$