Geometric Probability: Very Difficult Problems with Solutions

Solution:
The point [tex](X, Y, Z)[/tex] is uniformly distributed in the unit cube, whose volume is [tex]1[/tex], so probabilities are volumes. By the symmetry between [tex]X[/tex] and [tex]Z[/tex], the event [tex]X\lt Z[/tex] occupies half of the cube:
[tex]P(X\lt Z)=\frac{1}{2}[/tex]
The planes [tex]x=y[/tex], [tex]y=z[/tex] and [tex]x=z[/tex] cut the cube into [tex]6[/tex] congruent parts, one for each order of [tex]X[/tex], [tex]Y[/tex], [tex]Z[/tex] (equalities have zero volume), so each order has probability [tex]\frac{1}{6}[/tex]. The same follows by integration:
[tex]P(X\lt Y\lt Z)=\int_0^1\frac{z^2}{2}\,dz=\frac{1}{6}[/tex]
Since [tex]X\lt Y\lt Z[/tex] implies [tex]X\lt Z[/tex], the conditional probability is
[tex]P\left(X\le Y\le Z\mid X\lt Z\right)=\frac{P(X\lt Y\lt Z)}{P(X\lt Z)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{3}[/tex]
In other words, of the three equally likely orders with [tex]X\lt Z[/tex] ([tex]Y\lt X\lt Z[/tex], [tex]X\lt Y\lt Z[/tex], [tex]X\lt Z\lt Y[/tex]) exactly one has [tex]Y[/tex] between [tex]X[/tex] and [tex]Z[/tex]. The answer is not [tex]\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}[/tex], because the events [tex]X\lt Y[/tex] and [tex]Y\lt Z[/tex] are not independent.

Solution:
Let [tex]x[/tex] be the distance from the midpoint of the stick to the nearest boundary and [tex]\varphi[/tex] the angle between the stick and the boundaries. Then
[tex]0\le x\le 3 , \quad 0\le\varphi\le\frac{\pi}{2}[/tex]
and the point [tex](\varphi, x)[/tex] is uniformly distributed in a rectangle of area
[tex]\frac{\pi}{2}\times 3=\frac{3\pi}{2}[/tex]
Each half of the stick has length [tex]3[/tex], and its projection on the direction perpendicular to the boundaries is [tex]3\sin \varphi[/tex]. So the stick crosses the nearest boundary exactly when [tex]x\le 3\sin \varphi[/tex]; it cannot reach any other boundary, because the next one is at distance [tex]6-x\ge 3[/tex] from the midpoint. The area of the region [tex]x\le 3\sin \varphi[/tex] is
[tex]\int_0^{\frac{\pi}{2}}3\sin \varphi\,d\varphi=\left.-3\cos\varphi\right|_0^{\frac{\pi}{2}}=3[/tex]
The probability that the stick crosses a boundary is the ratio of the areas, [tex]\frac{3}{\frac{3\pi}{2}}=\frac{2}{\pi}[/tex], so the probability that it lies entirely on one stripe is
[tex]P=1-\frac{2}{\pi}\approx 0.363[/tex]
(This is Buffon's needle problem.)

Solution:
The answer does not depend on the size of the square, so take the square [tex]-1\le x\le 1[/tex], [tex]-1\le y\le 1[/tex] with center [tex]O(0, 0)[/tex]. The coordinate axes and the diagonals cut it into [tex]8[/tex] congruent triangles, and the condition is symmetric with respect to these lines, so it suffices to work in the triangle
[tex]T:\ 0\le y\le x\le 1 , \quad S_T=\frac{1}{2}[/tex]
In [tex]T[/tex] the nearest side is [tex]x=1[/tex], at distance [tex]1-x\ge 0[/tex]. Both sides of the inequality are nonnegative, so we may square:
[tex]\sqrt{x^2+y^2}\lt 1-x\iff x^2+y^2\lt 1-2x+x^2\iff x\lt\frac{1-y^2}{2}[/tex]
Together with [tex]x\ge y[/tex] this requires [tex]y\lt\frac{1-y^2}{2}[/tex], i.e. [tex]y^2+2y-1\lt 0[/tex], so [tex]0\le y\lt\sqrt{2}-1[/tex]. Let [tex]t=\sqrt{2}-1[/tex]; the favorable area in [tex]T[/tex] is
[tex]S=\int_0^{t}\left(\frac{1-y^2}{2}-y\right)dy=\frac{t}{2}-\frac{t^3}{6}-\frac{t^2}{2}=\frac{3t-t^3-3t^2}{6}[/tex]
Since [tex]t^2=3-2\sqrt{2}[/tex] and [tex]t^3=5\sqrt{2}-7[/tex], we get
[tex]S=\frac{3\sqrt{2}-3-5\sqrt{2}+7-9+6\sqrt{2}}{6}=\frac{4\sqrt{2}-5}{6}[/tex]
The probability is the ratio of the areas:
[tex]P=\frac{S}{S_T}=\frac{4\sqrt{2}-5}{3}\approx 0.22[/tex]