The total outcomes of dice are \(1,2,3,4,5,6\).
And, \(2,4\) and \(6\) are multiple of \(2\).
So, favourable cases \(=3\).
\(\therefore\) \(\text{Probability}=\frac{\text{Number}\ \text{of}\ \text{favourable}\ \text{cases}}{\text{Total}\ \text{cases}}\)
\(=\frac{3}{6}\)
\(=\frac{1}{2}\)
Hence, option \((C)\) is correct.