For the 3 inputs $A, B, C$ majority voting is performed if the number of $1’s$ is $\geq2$.
A |
B |
C |
Y |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
So $Y=A’BC+AB’C+ABC’+ABC$
after solving $Y$ using kmap we get $Y=AB+BC+CA$
Option (A) is correct.