Four digit numbers are formed using each of the digits 1, 2, ..., 8 only once. One number from them is picked up at random. The probability that the selected number contains unity is

$\frac{1}{2}$

Total number of four digit numbers formed with the digits 1, 2, 3, ....8 is $ {^8C}_4  × 4!$

Number of four digit numbers formed with the digits 1, 2,..... , 8 and containing unity as one of the digits is 

$ {^7C}_3  × 4!$

Hence, required probability = $\frac{^7C_3  × 4!}{^8C_4  × 4!}=\frac{1}{2}$