Pick a random point on the triangle with vertices (1,0,0), (0,1,0) and (0,0,1), what is the probability distribution of the x-coordinate? Diagram of the triangle. My attempt:
For $f(x) = 0$ for $x < 0$ or $x > 1$.
For $x = 0$, $f(x) = \sqrt{2}$ (the length of the $y-z$ side). For $x = 1$, $f(x) = 0$. So when $0 \leq x \leq 1$, $f(x) = -\sqrt{2}x + \sqrt{2}$.
However, I'm not convinced this is correct because from the diagram I feel as though $P(0.49\leq x \leq0.5)$ should be the same as $P(0.79\leq x \leq 0.8)$, for example.
 A: Your space triangle projects with constant area scalation to the triangle
$$T:=\bigl\{(x,y)\>\bigm|\>0\leq x\leq 1, \ 0\leq y\le 1-x\bigr\}$$
in the $(x,y)$-plane. The probability that a random point $(X,Y,Z)\in T$ satisfies $$X\leq x\qquad(0\leq x\leq 1)$$
therefore is given by
$$P(X\leq x)={\int_0^x(1-t)\>dt\over {\rm area}(T)}=2x-x^2\qquad(0\leq x\leq1)\ .$$
The probability density $f_X$ therefore is given by
$$f_X(x)=0 \quad \bigl(x\notin[0,1]\bigr), \qquad f_X(x)={d\over dx}P(X\leq x)=2-2x \quad (0\leq x\leq1)\ .$$
A: Let $A = (1,0,0)$, $B = (0,1,0)$ and $C = (0,0,1)$.
For any $\eta \in [0,1]$, let $B'$ and $C'$ be the intersection of
the plane $x = \eta$ with $AB$ and $AC$. It is easy to see $B' = (\eta,1-\eta,0)$ and $C' = (\eta,0,1-\eta)$.
In order for an uniformly sampled random point $P = (X,Y,Z)$ on $\triangle ABC$ to have $X \ge \eta$, $P$ need to belong to $\triangle AB'C'$.
Since $\triangle AB'C'$ is a scaled down version of $\triangle ABC$ for a factor $1-\eta$. Its area is $(1-\eta)^2$ of that of $\triangle ABC$.
Since $P$ is sampled uniformly over $\triangle ABC$, the probability for the event $X \ge \eta$ is proportional to the area of $\triangle AB'C'$. i.e.
$$\begin{align} & {\bf P}[X \ge \eta] = (1-\eta)^2\\
\implies &  {\bf P}[X \le \eta] = 1 - {\bf P}[X > \eta] = 1 - {\bf P}[X \ge \eta] = 2\eta - \eta\end{align}$$
The probability density function for the random variable $X$ is the derivative of the cumulative distribution function ${\bf P}[X \ge \eta]$. It is given by the expression:
$$\verb/PDF/(x) = \left.\frac{d}{d\eta}{\bf P}[X \ge \eta ]\right|_{\eta = x} = 2(1-x)$$
As a result, the distribution for $x$-coordinates is a triangular distribution.
The ratio for the probabilities for $0.49 \le X \le 0.5$ versus $0.79 \le X \le 0.8$ will be roughly $$(1-0.5) : (1-0.8) = 0.5 : 0.2 = 2.5$$ As one can see, this is very close to the exact ratio:
$${\bf P}[0.49 \le x \le 0.5] : {\bf P}[0.79 \le x \le 0.8] = 0.0101 : 0.0041 =  2.\overline{46341}$$
A: You ran into a problem similar to the Bertrand paradox and to quote wikipedia 'if, and only if, the method of random selection is specified, does the problem have a well-defined solution'. Uniform distribution of one variable (a point) over the two-dimentional plane the triangle lies on allows a projection to XY- or XZ-plane and from looking at that $PDF(x)=2-2x$ when $x \in [0,1]$ makes intuitive sense, while Your attempt is the answer if each coordinate is a variable with $Uniform(0,1)$ distribution. The question then would probably be 'pick a random point in a cube, what is pdf(x) of points that satisfy the equation of this triangle'.
