Propene reacts with $\mathrm{HBr}$ in the presence of an organic peroxide to give mainly

In the presence of peroxide the reaction proceeds by a radical chain. The $\mathrm{Br}{}^{\bullet }$ adds to the terminal carbon to give the more stable secondary radical, and this abstracts H from $\mathrm{HBr}$. The product is the anti-Markovnikov one, 1-bromopropane.