Geometric Probability: Problems with Solutions

Solution:
The number is a random point of the segment [tex]\left[0, 1\right][/tex] with all points equally likely, so the probability that it falls into a sub-segment equals the ratio of the length of the sub-segment to the length of the whole segment (geometric probability). Here the sub-segment is [tex]\left[0.30, 0.42\right][/tex] and the whole segment has length [tex]1[/tex]:
[tex]P=\frac{0.42-0.30}{1}=0.12[/tex]

Solution:
The point is chosen uniformly in the square, so the required probability equals the ratio of the area of the disk to the area of the square. Let the side of the square be [tex]a[/tex]. The diameter of the inscribed circle equals the side, so its radius is [tex]r=\frac{a}{2}[/tex]:
[tex]P=\frac{\pi r^2}{a^2}=\frac{\pi a^2}{4a^2}=\frac{\pi}{4}\approx 0.785[/tex]
The probability does not depend on the size of the square.

Solution:
Since [tex]x[/tex] is uniformly distributed in [tex]\left[4, 38\right][/tex], the probability equals the length of the favorable set divided by the length of the whole interval, [tex]38-4=34[/tex]. By the triangle inequality, each side must be less than the sum of the other two:
[tex]x\lt 8+15=23,\qquad 15\lt 8+x\iff x\gt 7,\qquad 8\lt 15+x[/tex]
The last inequality always holds, so the favorable set is the interval [tex]\left(7, 23\right)[/tex] of length [tex]16[/tex], which lies inside [tex]\left[4, 38\right][/tex]:
[tex]P=\frac{16}{34}=\frac{8}{17}[/tex]