Major Product of Alkene in Bromine Water
@orthocresol mentioned the main points in his comment but I would just like to expand a little more. The main electrophile in bromine water is $\ce{Br2}$ and this is what does the initial attack on the double bond, forming a cyclic bromonium ion. The bromonium ion is a strong electrophile and so it will react with any nucleophiles that are around. Since the water molecules will far outnumber bromide ions in the solution then the major nucleophile is water, even though it is not as good a nucleophile as bromide. The nucleophile attacks in an $\mathrm{S_N2}$ reaction, leading to a halohydrin product, after loss of a proton. This mechanism also explains the trans stereochemistry observed in the product.



As @orthocresol mentioned, the equilibrium constant for the formation of $\ce{BrOH}$ will be very small and so $\ce{BrOH}$ will be negligible as an electrophile. This is explained in a bit more detail in @Jan's answer here.