A five digit number is formed by the digits 1 ,   2 ,   3 ,   4 ,   5 with no digit…

No. of five digit numbers formed using the given digits $=5!$

We know that,

A number is divisible by $4$ if it's last two digits are divisible by $4$

Thus, the numbers should be of the form $\underline{} \underline{} \underline{} \underline{1} \underline{2}, \underline{} \underline{} \underline{} \underline{2} \underline{4}, \underline{} \underline{} \underline{} \underline{3} \underline{2}, \underline{} \underline{} \underline{} \underline{5} \underline{2}$

No. of such numbers $=4\times 3!$

Required probability $=\frac{4\times 3!}{5!}=\frac{1}{5}$