The digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 are written in random order to form a nine digit number. The probability that this number is divisible by 4, is

$\frac{2}{9}$

There are ${^9P_9}= 9!$ ways of arranging the given digits to form a nine digit number. So, total number of 9 digit numbers formed by the given digits = 9!

Out of these 9! numbers only those numbers are  divisible by 4 which have their last digits as even natural numbers and the numbers formed by their last two digits are divisible by 4.

The various possibilities of last two digits are :

12,  32,  52,   72,   92

24,  64,  84

16,  36,  56,  76,  96

28,  48,  68

This means that there are 16 ways of choosing the last two digits. Corresponding to each of these ways the remaining 7 digits can be arranged in 7! ways. Therefore, the total number of 9 digit numbers divisible by 4 is 16 × 7!

Hence, required probability $=\frac{16 × 7!}{9!}=\frac{2}{9}$