$$\begin{gathered} \text{Five digited numbers are formed using }\left\{0,2,4,5,7\right\} without repetition. One number is selected at random. \\ The probability that is divisible by 5 is \end{gathered}$$

Solution:

 $$\begin{gathered} \text{The number of five digit numbers formed using the digits 0,2,4,5,7 without repetition } \\ \text{n(S)=4×4!=96} \end{gathered}$$  

$$\begin{gathered} \text{A number is divisible by 5, if the unit digit is either zero or 5} \\ \text{The number of five digit numbers in which zero is in unit place is 4!=24} \\ \text{The number of five digit numbers in which five is in unit place is 3×3!=18} \\ \text{Hence the number of five digit numbers which are divisible by 5 is 24 +18=42} \\ \text{Hence the probability of the event is }\frac{42}{96}=\frac{7}{16} \end{gathered}$$