Given numbers are $1,2,3,4$
Possibilities for unit's place digit (either 1 or 3 )
$=2$
Possibilities for ten's digit $=3$
Possibilities for hundred's place digit $=2$
Possibilities for thousand' place's digit = 1
$\therefore$ Number of favourable outcomes
$=2 \times 3 \times 2 \times 1=12$
Number of numbers formed by $1,2,3,4$ (without repetitions) $=4 !$
$\therefore$ Required probability $=\frac{12}{4 \times 3 \times 2}=\frac{1}{2}$