Answer:
Option C
Explanation:
Given digits are 1,2,3,4
Possibilities for units 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
=$2 \times 3 \times2 \times 1=12$
Number of numbers formed by 1,2,3,4(without repetitions)=4!
$\therefore$ Required probability = $\frac{12}{4 \times 3 \times 2}=\frac{1}{2}$
