${\mathrm{CH}}_3–\mathrm{CH}={\mathrm{CH}}_2 + \mathrm{HI} \to \mathrm{X}$ Here product 'X' is

The outcome of the given reaction will follow Markownikoff's law since the peroxide effect only applies to HBr and is ineffective in HI. This can be explained with above reaction. ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HI} \to {\mathrm{CH}}_3-\underset{\text{ I}}{\underset{ \mid }{\mathrm{C}}}\mathrm{H}-{\mathrm{CH}}_3$