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 $=\frac{12}{4 \times 3 \times 2}=\frac{1}{2}$