4. Given a regular tetrahedron $P-ABC$ with a base edge length of 1 and a height of $\sqrt{2}$, then its inscribed sphere radius is $\qquad$
As shown in the figure, let the center of the base be $O$, and connect $BO$ to intersect $AC$ at $D$, then $OD=\frac{\sqrt{3}}{6} \Rightarrow PD=\sqrt{\left(\frac{\sqrt{3}}{6}\right)^{2}+2}=\frac{5 \sqrt{3}}{6}$. Let the incenter of the regular tetrahedron be $I$, and the radius be $r$. Draw $IE \perp PD$, then $\triangle PEI \sim \triangle POD \Rightarrow \frac{EI}{OD}=\frac{PI}{PD}$ $\Rightarrow \frac{r}{\frac{\sqrt{3}}{6}}=\frac{\sqrt{2}-r}{\frac{5 \sqrt{3}}{6}} \Rightarrow 5r=\sqrt{2}-r \Rightarrow r=\frac{\sqrt{2}}{6}$.