Geometric Probability: Difficult Problems with Solutions

Solution:
Let [tex]x[/tex] and [tex]y[/tex] be the minutes after [tex]10{:}00[/tex] at which the bus and the passenger arrive. The point [tex]\left(x, y\right)[/tex] is uniformly distributed in the rectangle [tex]0\le x\le 5[/tex], [tex]0\le y\le 15[/tex], whose area is
[tex]S=5\times 15=75[/tex]
The bus leaves at the moment [tex]x+5[/tex]. If the passenger comes earlier, he waits, so he catches the bus exactly when [tex]y\le x+5[/tex]. For a fixed [tex]x[/tex] the favorable values of [tex]y[/tex] form a segment of length [tex]x+5[/tex], so the favorable region is a trapezoid with parallel sides [tex]5[/tex] and [tex]10[/tex] and height [tex]5[/tex]:
[tex]S_1=\frac{5+10}{2}\times 5=37.5,\qquad P=\frac{S_1}{S}=\frac{37.5}{75}=\frac{1}{2}[/tex]

Solution:
Measure time in minutes after [tex]10{:}00[/tex] and let [tex]x[/tex] and [tex]y[/tex] be the arrival times of the two people. The point [tex]\left(x, y\right)[/tex] is uniformly distributed in the square [tex]0\le x\le 60[/tex], [tex]0\le y\le 60[/tex] of area [tex]60^2=3600[/tex]. The two meet exactly when their arrival times differ by at most [tex]5[/tex] minutes:
[tex]|x-y|\le 5[/tex]
The complementary region [tex]|x-y|\gt 5[/tex] consists of two right isosceles triangles with legs [tex]60-5=55[/tex]; their total area is
[tex]2\times \frac{55^2}{2}=3025\ \Rightarrow\ P=\frac{3600-3025}{3600}=\frac{575}{3600}=\frac{23}{144}[/tex]

Solution:
Let [tex]x[/tex] and [tex]y[/tex] be the arrival times (in hours) of the first and of the second ship. The point [tex]\left(x, y\right)[/tex] is uniformly distributed in the square [tex]0\le x\le 24[/tex], [tex]0\le y\le 24[/tex] of area [tex]24^2=576[/tex], so probabilities are ratios of areas. A ship has to wait if the second ship arrives while the first one is still at the berth, or the first arrives while the second one is there:
[tex]x\le y\lt x+2\text{ or }y\le x\lt y+1[/tex]
Nobody waits when [tex]y\ge x+2[/tex] or [tex]x\ge y+1[/tex]. These are two right isosceles triangles in the square with legs [tex]24-2=22[/tex] and [tex]24-1=23[/tex]; their total area is
[tex]\frac{22^2}{2}+\frac{23^2}{2}=242+264.5=506.5[/tex]
[tex]P=1-\frac{506.5}{576}=\frac{69.5}{576}=\frac{139}{1152}[/tex]

Solution:
The rectangle consists of the points with [tex]-2\le x\le 2[/tex] and [tex]-0.5\le y\le 1[/tex], so its area is
[tex]S_{ABCD}=4\times 1.5=6[/tex]
The probability equals the area of the part of the disk inside the rectangle divided by [tex]6[/tex]. Every point of the disk satisfies [tex]|x|\le 1[/tex] and [tex]y\le 1[/tex], so only the line [tex]y=-0.5[/tex] cuts off a part of the disk: a circular segment whose chord is at distance [tex]\frac{1}{2}[/tex] from the center. If its central angle is [tex]2\theta[/tex], then
[tex]\cos\theta=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{3}\Rightarrow 2\theta=\frac{2\pi}{3}[/tex]
The area of the segment is the area of the sector minus the area of the triangle:
[tex]S_1=\frac{1}{2}\times 1^2\times \frac{2\pi}{3}-\frac{1}{2}\times 1^2\times \sin \frac{2\pi}{3}=\frac{\pi}{3}-\frac{\sqrt{3}}{4}[/tex]
So the part of the disk inside the rectangle has area
[tex]S=\pi-\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)=\frac{2\pi}{3}+\frac{\sqrt{3}}{4}[/tex]
Hence
[tex]P=\frac{S}{6}=\frac{1}{6}\left(\frac{2\pi}{3}+\frac{\sqrt{3}}{4}\right)=\frac{\pi}{9}+\frac{\sqrt{3}}{24}\approx 0.421[/tex]

Solution:
The point [tex](a, b)[/tex] is a random point of a rectangle of area [tex]1\times 2=2[/tex], so the probability equals the area of the favorable region divided by [tex]2[/tex].
Since [tex]a\gt 0[/tex], the equation is quadratic, and it has real roots exactly when its discriminant is nonnegative:
[tex]D=b^2-4a\ge 0\iff a\le\frac{b^2}{4}[/tex]
For [tex]-1\lt b\lt 1[/tex] we have [tex]\frac{b^2}{4}\lt 1[/tex], so the favorable region is the part of the rectangle between the line [tex]a=0[/tex] and the parabola [tex]a=\frac{b^2}{4}[/tex]. Its area is
[tex]S=\int_{-1}^{1}\frac{b^2}{4}\,db=\left.\frac{b^3}{12}\right|_{-1}^{1}=\frac{1}{12}+\frac{1}{12}=\frac{1}{6}[/tex]
Hence
[tex]P=\frac{S}{2}=\frac{1}{12}[/tex]