A point M is thrown at random into the square with vertices…

A point $M$ is thrown at random into the square with vertices $(0,0),(0,1),(1,0),(1,1)$; let $(u,v)$ be its coordinates. The probability that the point falls into a region of the square is proportional to the area of the region.
$1^{\circ}$ Prove that for $0\lt x,y\lt1$: $\ P(u\lt x,\ v\lt y)=P(u\lt x)\,P(v\lt y)=xy$.
$2^{\circ}$ For $0\lt z\lt1$ find: 1) $P(|u-v|\lt z)$; 2) $P(uv\lt z)$; 3) $P(\min(u,v)\lt z)$; 4) $P(\max(u,v)\lt z)$; 5) $P\big(\tfrac12(u+v)\lt z\big)$.

Step 1: $1^{\circ}$ — independence of the coordinates

$\{u\lt x,\ v\lt y\}$ is the rectangle $[0,x)\times[0,y)$ of area $xy$. Also $\{u\lt x\}$ is the strip $[0,x)\times[0,1]$ of area $x$, and $\{v\lt y\}$ has area $y$. So

\[ \boxed{P(u\lt x,\ v\lt y) = xy = P(u\lt x)\,P(v\lt y).} \]

The events $\{u\lt x\}$ and $\{v\lt y\}$ are independent for all $x,y$: $u$ and $v$ are independent random variables, each uniform on $(0,1)$.

Step 2: 1) $P(|u-v|\lt z)$

The band around the diagonal; the complement consists of two corner triangles with legs $1-z$ (compare Problem 71):

\[ P(|u-v|\lt z) = 1-(1-z)^2 = 2z-z^2. \]

Step 3: 2) $P(uv\lt z)$

The complement $uv\ge z$ lies above the hyperbola $v=z/u$, which meets $v=1$ at $u=z$:

\[ P(uv\ge z) = \int_z^1\Big(1-\frac zu\Big)du = (1-z) + z\ln z, \]

\[ P(uv\lt z) = 1-(1-z)-z\ln z = z - z\ln z = z(1-\ln z). \]

Step 4: 3) $P(\min(u,v)\lt z)$

The minimum is $\ge z$ exactly when both coordinates are $\ge z$: the square $[z,1]^2$ of area $(1-z)^2$. So

\[ P(\min(u,v)\lt z) = 1-(1-z)^2 = 2z-z^2. \]

(The same number as in 1), although the regions are different.)

Step 5: 4) $P(\max(u,v)\lt z)$

The maximum is $\lt z$ exactly when both coordinates are $\lt z$: the square $[0,z)^2$:

\[ P(\max(u,v)\lt z) = z^2. \]

Step 6: 5) $P\big(\tfrac12(u+v)\lt z\big)$

The condition is $u+v\lt 2z$, the region below the line $u+v=2z$.

• If $z\le\frac12$ ($2z\le1$): a triangle with legs $2z$, area $\frac12(2z)^2=2z^2$.
• If $z\gt\frac12$: the complement is a triangle in the upper corner with legs $2-2z$, area $\frac12(2-2z)^2 = 2(1-z)^2$.

\[ P\big(\tfrac12(u+v)\lt z\big) = \begin{cases} 2z^2, & 0\lt z\le\frac12,\\[2pt] 1-2(1-z)^2, & \frac12\lt z\lt1. \end{cases} \]

Step 7: Numerical check at $z=\frac12$