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Given a square $S$ with size $1 \times 1$. Two randomly selected points $A$ and $B$ are inside the square. Let $U$ be a square with diagonal $AB$. How to find out the probability $P$($U$ is inside $S$). example

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  • Define a coordinate system with origin at point W. Then express the coordinates of $L$ and $T$ as functions of the coordinates of $A$ and $B$. That will allow you to find what conditions the coordinates of $A$ and $B$ must satisfy in order for the coordinates of $L$ and $T$ to all be between $0$ and $1$. And then you should be able to relatively easily find the probability that that condition is satisfied. – Keshav Srinivasan Oct 10 '13 at 17:10

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Assume that the square $S$ is $(0,1)\times(0,1)$ and that the selected points $A$ and $B$ are $(X,Y)$ and $(Z,T)$. Then, $X$, $Y$, $Z$ and $T$ are i.i.d. and uniform on $(0,1)$ and $$ [U\subset S]=C\cap D, $$ where $$ C=[2W\lt X+Y+Z+T],\qquad D=[X+Y+Z+T\lt2+2V]. $$ and $$ V=\min(X,Y,Z,T),\qquad W=\max(X,Y,Z,T). $$ It remains to compute $P[C\cap D]$.

The symmetry $$ (X,Y,Z,T,V,W)\to(1-X,1-Y,1-Z,1-T,1-W,1-V) $$ exchanges $C$ and $D$ hence it leaves the event $[U\subset S]$ invariant and one has $P[C]=P[D]$. Since $[2W\gt2+2V]=\varnothing$, $C^c\cap D^c=\varnothing$, that is, $C\cup D=\Omega$ hence $$ P[C\cap D]=2P[C]-1. $$ It remains to compute $P[C]$.

Assume without loss of generality that $W=T$, then, conditionally on $[T=t]$, $X$, $Y$ and $Z$ are i.i.d. and uniformly distributed on $(0,t)$. Thus, $P[C\mid W=T=t]=1-|\Delta_t|/t^3$, where $$ \Delta_t=\{(x,y,z)\in(0,t)^3\mid x+y+z\lt t\}. $$ By scaling, $P[C\mid W=T=t]=1-|\Delta_1|$. This does not depend on $t$ hence $$ P[C]=1-|\Delta_1|. $$ It remains to compute $|\Delta_1|$.

If $X$, $Y$ and $Z$ are once again i.i.d. and uniform on $(0,1)$, the distribution of $X+Y+Z$ has CDF $P[X+Y+Z\lt u]=\frac16u^3$ for every $u$ in $(0,1)$ (for higher values of $u$, the formula changes). In particular, $P[X+Y+Z\lt1]=|\Delta_1|=\tfrac16$.

Finally, $P[C]=\tfrac56$ hence $$ P[U\subset S]=\tfrac23. $$

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