Triangle Probablity Distribution Given the following probability distribution +P(X>0.75)$. I've found that $$P(X<-0.75)+P(X>0.75)=2(\frac{(1-0.75)\cdot h}{2}),$$ where $h$ is the height of the shaded triangle. How do we get the height? Answer is $\frac{1}{16}$ **Edit 1:** Since this is a probability distribution, the entire area $= 1$. Area of entire triangle: $\frac{(1-(-1))\cdot H}{2} = 1$ $H = 1$ We split the entire triangle in half vertically down the middle at $0$ so we get two triangles which have all 3 corner angles equal to its respective shaded triangles contained within. Now we use the property of similar triangles where the side-lengths of the shaded triangles are proportionate to its parent triangle. $h = \frac{1-0.75}{1-0}\cdot H$ $h = \frac{1}{4}\cdot 1$ $h = \frac{1}{4}$ Thus, $P(X<-0.75)+P(X>0.75)$ $=2(\frac{(1-0.75)\cdot \frac{1}{4}}{2})$ $=\frac{1}{16}$

A probability distribution integrates to $1$, i.e. the area of the big triangle must be $1$. What must its height then be? Now use similar triangles to compute the height of the smaller triangles.