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

$\frac{1}{2}$

The 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 $=\frac{^7C_3 × 4 !}{^8C_4 × 4 !}=\frac{1}{2}$