In a box, there are 10 identical balls, numbered from 1 to 10. The number of each ball corresponds to one of the points on the figure, which divide the circumference into 10 equal parts. In the following items, consider that the balls are drawn at random, one by one and without replacement.

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a) If two balls are drawn, what is the probability that the segment determined by the corresponding points is a diameter of the circle?

b) If three balls are drawn, what is the probability that the corresponding points are vertices of a right triangle?

An inscribed angle in a circle is a right angle if and only if the corresponding arc is a semicircle.
c) If four balls are drawn, what is the probability that the corresponding points are vertices of a rectangle?
Solution

a) 1st solution: The multiplication principle shows that the number of ways to draw two balls, one at a time, is $10 \times 9=90$. Of these draws, there are ten for which the segment determined by the points drawn is a diameter, namely, $(1,6),(2,7),(3,8),(4,9),(5,10),(6,1),(7,2),(8,3),(9,4)$, and $(10,5)$. Therefore, the required probability is $\frac{10}{90}=\frac{1}{9}$.

2nd solution: A ball is drawn at random. Of the nine possibilities for drawing another ball, only one will, together with the first, determine a diameter. Therefore, the probability of drawing two balls that determine a diameter is $\frac{1}{9}$.

3rd solution: It is possible to draw two balls in $\binom{10}{2}=45$ different ways. Of these draws, there are five that determine diameters; thus, the sought probability is $\frac{5}{45}=\frac{1}{9}$.

b) 1st solution: The multiplication principle shows that the number of ways to draw three balls, one at a time, is $10 \times 9 \times 8=720$. For a draw to determine a right triangle, it must contain two balls $a$ and $b$ that determine a diameter and a third ball $x$ distinct from these two. Arranging these three balls in the $3!=6$ possible ways, we see that there are six draws that consist of these balls. Since there are five pairs of balls that determine a diameter and the extra ball can be chosen in eight different ways, the number of draws that determine an inscribed right triangle is $6 \times 5 \times 8=240$. Therefore, the sought probability is $\frac{240}{720}=\frac{1}{3}$.

2nd solution: Once three balls are drawn, we can form three groups of two balls from them. We observe that if one of these groups determines a diameter, then this cannot happen for the other two groups. Since each group of two balls has a probability of $\frac{1}{9}$ of determining a diameter, the sought probability is then $\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{1}{3}$.

3rd solution: There are $\binom{10}{3}=120$ ways to choose three balls, that is, there are 120 inscribed triangles with vertices at the vertices of the decagon. On the other hand, each diameter determines eight inscribed right triangles, for a total of $5 \times 8=40$; that is, there are 40 choices of three balls that determine inscribed right triangles. The sought probability is then $\frac{40}{120}=\frac{1}{3}$.

c) 1st solution: The number of draws of four balls is $10 \times 9 \times 8 \times 7$ and each of these draws determines an inscribed quadrilateral. On the other hand, the balls of a draw that determines an inscribed rectangle must determine two diameters. There are ten choices for the first ball of such a draw and the diametrically opposite ball can then appear in any of the three subsequent positions; the other two balls can then be chosen in eight different ways, corresponding to the four diameters not yet determined. Thus, the draws that determine a right triangle are in number of $10 \times 3 \times 8$ and the sought probability is then $\frac{10 \times 3 \times 8}{10 \times 9 \times 8 \times 7}=\frac{1}{21}$.

2nd solution: For the four balls drawn to determine a rectangle, the first three must determine a right triangle, which happens with probability $\frac{1}{3}$; once this is done, there is a unique choice for the fourth ball among the seven remaining. Therefore, the sought probability is $\frac{1}{3} \times \frac{1}{7}=\frac{1}{21}$.

3rd solution: There are $\binom{10}{4}=210$ ways to choose four balls, that is, there are 210 inscribed quadrilaterals with vertices at the vertices of the decagon. On the other hand, an inscribed rectangle is determined by two diameters, that is, there are $\binom{5}{2}=10$ inscribed rectangles, corresponding to ten choices of four balls. Therefore, the sought probability is $\frac{10}{210}=\frac{1}{21}$.