Reaction of HBr with propene in the presence of peroxide gives:

Reaction of $\mathrm{HBr}$ with propene in the presence of peroxide gives $n$-propyl bromide. This addition reaction is an example of anti-Markownikoffs addition reaction.
Related Theory Peroxide undergoes fission to give free radicals. $\mathrm{R}-\mathrm{O}-\mathrm{O}-\mathrm{R} \longrightarrow2 \mathrm{R}-\dot{\mathrm{O}} \mathrm{HBr}$ combines with free radical to form bromine free radical. R-\dot{O}+$\mathrm{HBr} \longrightarrow \mathrm{R}-\mathrm{OH}+\dot{\mathrm{B}}\dot{B}$ attacks the double bond of the alkene to form a more stable free radical.$<img src="/api/proxy-image?url=https$ \\ More stable free radical attacks the $\mathrm{HBr}$. \begin{aligned}\mathrm{CH}_{3} \dot{\mathrm{C}} \mathrm{CHH} }_{2} \mathrm{Br}+\mathrm{HBr} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}}_{2} \mathrm{CH}_{2} \mathrm{Br}+\dot{\mathrm{B}}\\ & \text{ n-propylbromide } \end{aligned}\dot{B}$r+\dot{B} r \longrightarrow B r_2$