The major product formed in the following reaction is:
$$\text{CH}_2\text{CH} = \text{CHCH(CH}_3)_2 \xrightarrow{\text{HBr}}$$

To determine the major product formed in the reaction of $$CH_2=CHCH(CH_3)_2$$ with $$HBr$$, we apply Markovnikov’s rule while considering the possibility of carbocation rearrangement.

The reaction proceeds through an electrophilic addition mechanism. In the first step, the proton ($$H^+$$) from $$HBr$$ adds to the double bond. According to Markovnikov’s rule, the proton attaches to the terminal carbon atom, generating the more stable secondary carbocation.

The protonation step is represented as

$$CH_2=CH-CH(CH_3)_2 \xrightarrow{H^+} CH_3-CH^+-CH(CH_3)_2.$$

The resulting secondary carbocation is adjacent to a carbon atom bearing a hydrogen atom. Consequently, a 1,2-hydride shift occurs, producing a more stable tertiary carbocation.

The rearrangement is represented as

$$CH_3-CH^+-CH(CH_3)_2 \longrightarrow CH_3-CH_2-C^+(CH_3)_2.$$

Finally, the bromide ion ($$Br^-$$) attacks the tertiary carbocation to give the final product,

$$CH_3CH_2C(Br)(CH_3)_2.$$

Thus, owing to the formation of the more stable tertiary carbocation through hydride shift, the major product of the reaction is

$$\boxed{CH_3CH_2C(Br)(CH_3)_2},$$

which corresponds to option D.