smc

Clearly $BC=5$ and $AC=5\sqrt{3}$. Choose a $P^{\prime }$ and get a corresponding $D^{\prime }$ such that $B{D}^{\prime }=5\sqrt{2}$ and $C{D}^{\prime }=5$. For $BD>5\sqrt{2}$ we need $CD>5$, creating an isosceles right triangle with hypotenuse $5\sqrt{2}$ . Thus the point $P$ may only lie in the triangle $AB{D}^{\prime }$. The probability of it doing so is the ratio of areas of $AB{D}^{\prime }$ to $ABC$, or equivalently, the ratio of $A{D}^{\prime }$ to $AC$ because the triangles have identical altitudes when taking $A{D}^{\prime }$ and $AC$ as bases. This ratio is equal to $\frac{AC-C{D}^{\prime }}{AC}=1-\frac{C{D}^{\prime }}{AC}=1-\frac{5}{5\sqrt{3}}=1-\frac{\sqrt{3}}{3}=\frac{3-\sqrt{3}}{3}$. Thus the answer is $\boxed{\frac{3-\sqrt{3}}{3}}$.