784 Combinatorics and Probability
D={(x, y)| 0 ≤x≤ 1 , 0 ≤y≤ 1 }.In order for the two people to meet, their arrival time must lie inside the region
Df={
(x, y)||x−y|≤1
4
}
.
The desired probability is the ratio of the area of this region to the area of the square.
The complement of the region consists of two isosceles right triangles with legs equal
to^34 , and hence of areas^12 (^34 )^2. We obtain for the desired probability
1 − 2 ·
1
2
·
(
3
4
) 2
=
7
16
≈ 0. 44.
(B.V. Gnedenko)930.The set of possible events is modeled by the square[ 0 , 24 ]×[ 0 , 24 ]. It is, however,
better to identify the 0th and the 24th hours, thus obtaining a square with opposite sides
identified, an object that in mathematics is called a torus (which is, in fact, the Cartesian
product of two circles. The favorable region is outside a band of fixed thickness along
the curvex=yon the torus as depicted in Figure 110. On the square model this region
is obtained by removing the points(x, y)with|x−y|≤1 together with those for which
|x−y− 1 |≤1 and|x−y+ 1 |≤1. The required probability is the ratio of the area of
the favorable region to the area of the square, and is
P=
242 − 2 · 24
242
=
11
12
≈ 0. 917.
Figure 110931.We assume that the circle of the problem is the unit circle centered at the originO.
The space of all possible choices of three pointsP 1 ,P 2 ,P 3 is the product of three circles;
the volume of this space is 2π× 2 π× 2 π= 8 π^3.