Answer: B
In the ionic path the proton adds to C1 to give a secondary cation, which undergoes a 1,2-hydride shift to the tertiary cation. The bromide then gives 2-bromo-2-methylbutane. In the radical path BrX∙ adds to C1 and the resulting secondary radical takes H from HBr without rearrangement, so 1-bromo-3-methylbutane forms.