Probability of Airplane Parking at the same Gate

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 curve $x = y$ on the torus as depicted in the Figure. 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 $P = \dfrac{24^2 - 2 \cdot 24}{24^2} = \dfrac{11}{12} \approx 0.917$

Note that this solution is taken exactly it is given from the book mentioned above.

Here is a picture of the joint distribution of the plane arrival times within a 24-hour period. The planes cannot arrive within 1 hour of each other.

Therefore the probability is $\frac{23^2}{24^2}$

so that $p+q=23^2+24^2=1105$ .

The airplanes can both park if $| t_1-t_2 | >=1h$ , so if the first time is known, 22 out of 24 parts of the day will work for the second, hence the solution is just 23.