A 4 digit number made of digits 1, 2, 3, 4, 5 is written down at random without repetition. The probability that the number so formed is divisible by 6 is

The given digits are $1,2,3,4,5$

The number of four digited numbers from the given five digits without repetition is $n\left(S\right)=5_{P_4}=5!=120$

Any four digit number is divisible by 6, if it is even number and sum of the digits must be divisible by 3. 

Unit digit must be filled by either 2 or 4

Suppose that 2 is in unit place, the remaining three digits can be filled with $1,4,5$ in order to make sum of the digits is divisible by 3. this can be done in $3!$ ways

Suppose that 4 is in unit place, the remaining three digits can be filled with $1,2,5$ in order to make sum of the digits is divisible by 3, this can be done in $3!$ ways

Therefore, the number of four digited numbers which are divisible by 6 is $2\left(3!\right)=12$

Therefore, the required probability is $\frac{12}{120}=\boxed{\frac{1}{10}}$