A metal coin of diameter 2r< a is thrown onto an infinite…

A metal coin of diameter $2r\lt a$ is thrown onto an infinite chessboard with squares of side $a$. Find the probability that:
$1^{\circ}$ the whole coin falls inside one of the squares;
$2^{\circ}$ the coin intersects exactly one side of a square.

Step 1: Reduce to one square

The board repeats periodically, so we only need to know where the centre $O$ of the coin lies inside the square that contains it. "Thrown at random" means $O$ is uniformly distributed over that square of area $a^2$.

The coin (radius $r$) meets a side of the square exactly when the distance from $O$ to that side is less than $r$.

Step 2: $1^{\circ}$ — the coin inside one square

The centre must be at distance at least $r$ from all four sides: it lies in the inner square of side $a-2r$ (green).

\[ \boxed{P_1 = \frac{(a-2r)^2}{a^2} = \left(1-\frac{2r}{a}\right)^2.} \]

Step 3: $2^{\circ}$ — the coin crosses exactly one side

The centre is within $r$ of exactly one side: along each side there is a strip of width $r$ and length $a-2r$ (orange), excluding the corner squares where it would be close to two sides. Four such strips:

\[ \boxed{P_2 = \frac{4r(a-2r)}{a^2}.} \]

Step 4: Check — the whole square

The remaining region consists of $4$ corner squares $r\times r$ (red), where the coin crosses two sides (near a corner of the board). Adding up:

\[ (a-2r)^2 + 4r(a-2r) + 4r^2 = a^2. \ \checkmark \]

(Since $2r\lt a$, the coin can never meet two parallel sides.)

Step 5: Example

Squares of side $a=5$ cm and a coin of diameter $2r=2$ cm ($r=1$):

\[ P_1 = \left(\tfrac35\right)^2 = 0.36, \qquad P_2 = \frac{4\cdot1\cdot3}{25}=0.48, \qquad P(\text{two sides}) = \frac{4}{25}=0.16. \]