The rearrangement which most likely occurs in the following reaction can be described as a HBr
(CH3)2CHCH=CH2 →

a. hydride shift from C-2 to C-1.
b. hydride shift from C-3 to C-2.
c. proton shift from C-2 to C-1.
d. methyl group shift from C-3 to C-2.

b. hydride shift from C-3 to C-2.

Explanation:

Markovnikov's rule states that *in the addition of a protic acid HX or other polar reagent to an asymmetric alkene, the acid hydrogen (H) or electropositive part gets attached to the carbon with more hydrogen substituents, and the halide (X) group or electronegative part gets attached to the carbon with more alkyl substituents* (wikipedia).

This rule implies that the hydrogen of HBr will be attached to C-1 and the carbocation will be on C-2. Remember that the order of stability of carbocations is tertiary > secondary > primary > methyl. A hydride shift can yield a tertiary carbocation.

C-3 is a tertiary carbon atom. If the hydride on carbon 3 shifts to carbon 2, a tertiary and more stable carbocation is formed. This accounts for the major product in the reaction.