Propene reacts with $\mathrm{Br}{}_2$ in water. The major product is

Br adds first to give the bromonium ion. Water attacks the carbon that carries more positive charge, the more substituted carbon. Br ends on $\mathrm{CH}{}_2$ and OH on $\mathrm{CH}$, giving $\mathrm{BrCH}{}_2\mathrm{CH}(\mathrm{OH})\mathrm{CH}{}_3$.