What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP$? 
A point $P$ is chosen at random in the interior of the equilateral
  triangle $ABC$.What is the probability that $\triangle ABP$ has a
  greater area than each of $\triangle ACP$ and $\triangle BCP$?


Since the three triangles $ABP,ACP$ and $BCP$ have equal bases,their areas are proportional to their length of altitudes.
But i dont know how to solve further and find the required probability.Please help me.Thanks. 

There are six equally likely outcomes:
$ABP>BCP>ACP$
$ABP>ACP>BCP$
$BCP>ACP>ABP$
$BCP>ABP>ACP$
$ACP>BCP>ABP$
$ACP>ABP>BCP$
Each of them are mutually exclusive and their sum is $1$. Hence $BCP$ being the largest has probability $1\over3$. (The equal area case has probability $0$ and can be ignored because we have a dense space, that is, $>$ sign has same probability as $\geq$ sign.)
