Mathematics 12.

In the following we will talking about probabilistic experiments where the probability of events are determined using a geometric method.

Let us take, for example, a spinning top; spin it, and when it stops, a fixed arrow will point at a random point on the circumference.

If all the spinning tops presented on Figure 1 have one third painted red, blue and green, respectively, we expect that all colors have \latex{\frac{1}{3}} chance to be pointed at by the arrow when the spinning stops, no matter which arrangement is used.

Thus the probability of the individual colors does not depend on how the sectors are arranged, but only on the size of the related arc on the circumference of the spinning top, or the central angle, or the area of the sector.

It sounds natural that if we use the spinning top presented on Figure 2, we have \latex{\frac{1}{3}} chance to spin red, \latex{\frac{1}{6}} chance to spin green and \latex{\frac{1}{2}} chance to spin blue. 

Consider a target which is hit by a random shot. An elementary event is a hit of an arbitrary point of the target, and the points of the table make up the event space. In contrast with the previous experiments, this event space is infinite. The probability that the shot arrives at the target is \latex{ 1 }, although the probability of hitting a given point, for example the bull's eye, is \latex{ 0 }. This is an event whose probability is \latex{ 0 }, although it is not impossible.

Stating that the shot hits the target at random or follows a uniform distribution means that the probability of the shot hitting a given section of the target is proportional to the area of that section, 

\latex{P(A)=c\times T(A),}

where \latex{ c } is the said constant ratio. If \latex{ H } stands for the certain event which corresponds to the whole of the target, then

\latex{1=P(H)=c\times T(H),}

from which

Using this for an event \latex{ A } of our choice,

The probability of an event \latex{ A } is the ratio of the area which corresponds to \latex{ A } and the whole area (Figure 3). Depending on the experiment, length or angle can also be used instead of area.

On the first target, the colored square takes up half of the target's area, therefore the probability that the shot hits the colored square is \latex{\frac{1}{2}.}

The second target has \latex{ 4 } colored squares each of whose area is \latex{\frac{1}{8}} part of the large square, therefore in total, the colored area is the half of the area of the square. The probability of a random shot hitting the colored part is \latex{\frac{1}{2}.}

If the side of the large square is divided into k equal parts, then there
are \latex{k^{2}} colored squares whose area is \latex{\frac{1}{2k^{2} }} times that of the large square, thus the sum of them is the half of the whole area.

Therefore the proportion of the color area is the same in the case of every square, which is not surprising since the large square can be divided into small ones, and half of their area will be that of the colored square. It is interesting to note that although the area of the colored squares is the same for consecutive targets, their circumference is increasing beyond all limits.

If random shots are fired at the targets presented above and we check what is the relative frequency of the shot hitting a colored area, the result will be approximately the same in case of all targets.

This verifies our hypothesis that the probability of hitting the observed area depends only on the size of that area, but not on its position.

Therefore for each target, a shot fired randomly at the target will hit a colored square with a probability of \latex{\frac{1}{2}.}

First of all let us examine which points are those within the hexagon for which the normals to the sides \latex{ AB } and \latex{ ED } will end in internal points of these sides. Since the square \latex{ ABDE } is a rectangle, the points of this rectangle are exactly those satisfying the above criteria. Similarly, for the internal points of the rectangle \latex{ BCEF }, the endpoints of the normals to the sides \latex{ BC } and \latex{ EF }will be internal points of said sides and for the internal points of the rectangle \latex{ FACD }, the endpoints of the normals to the sides \latex{ FA } and \latex{ CD } will be internal points of said sides. The points for which the above criterion is fulfilled for all sides of the hexagon are in the intersection of the three rectangles. Due to the rotational symmetry of the hexagon, this intersection is also a regular hexagon. (Figure 4)

To determine the probability in question, we must know the proportion of the area of this smaller hexagon compared to that of the original one. By drawing the lines presented on Figure 5 into the small hexagon we can see that \latex{ BGH } and \latex{ GHO } are congruent regular triangles, and all triangles at the vertices have their counterpart within the small hexagon. Comparing the angles and lengths of the sides, it is easy to see that the area of the triangle \latex{ BHC } equals to that of the triangle \latex{ BGH }. Overall, the area of the large hexagon is \latex{ 18 } times that of the triangle \latex{ BGH }, while the area of the small hexagon is \latex{ 6 } times that of the same triangle.

Therefore the probability sought is  \latex{\frac{6}{18}=\frac{1}{3}.}

Let us demonstrate the time span between \latex{8^{30}} and \latex{9^{00}}  by a line segment, and let us mark the arrival times of the buses there. By marking those parts on this segment where no more than \latex{ 5 } minutes waiting is necessary after arrival, we can see that these make up \latex{ 15 } minutes out of the total of \latex{ 30 } minutes, therefore the probability in question is \latex{\frac{15}{30}=\frac{1}{2}.} (Figure 6)

First of all check out what the odds are for winning by choosing spinning top \latex{ A } for the first time and \latex{ B } for the second. Take a unit square and divide its sides according to spinning top \latex{ A } and spinning top \latex{ B } respectively as seen in Figure 7.

The figure tells us that the chance of spinning blue + red = purple, using geometric probability, is:

\latex{\frac{1}{4}\times \frac{1}{2}+\frac{1}{4}\times \frac{1}{6}=\frac{4}{24}=\frac{1}{6}.}

It can be seen as well that the order of spinning the two spinning tops is irrelevant.

Similarly, we can see that by choosing spinning top \latex{ A } twice subsequently will result in the probability to spin purple being \latex{\frac{1}{8}} (Figure 8), while choosing spinning top \latex{ B } twice, the probability will be  \latex{\frac{1}{6}}  (Figure 9).

So either two different spinning tops should be chosen, or \latex{ B } for both spins.

Note: Find what is worth choosing if we can select the second spinning top after observing the result of the first spin.

Susie arrives \latex{ x } hours after \latex{ 6 } pm and Betty arrives \latex{ y } hours after \latex{ 6 } pm. Since \latex{0\leq x\leq 1} and \latex{0\leq y\leq 1} , the event space is a unit square with a vertex at \latex{ O }. This means that for every point of the square there are two times when Susie and Betty arrive at the shop, and vice versa, for any pair of arrival times there is a point of the square that belongs to it.

If Susie arrives earlier, that is, \latex{x\leq y,} then the condition of meeting is

If Betty arrives sooner, or in other words, \latex{y\leq x,} the condition of meeting is:

Color the points corresponding to meeting red; one part consists of points satisfying the inequalities \latex{x\leq y} and \latex{x+\frac{1}{6}\geq y,} the other part consists of points satisfying the inequalities \latex{y\leq x} and \latex{y\geq x-\frac{1}{6}.} (Figure 10)

The area of red points in the unit square is

\latex{1-\left(\frac{5}{2} \right)^{2}=\frac{11}{36},}

since the area of the square is \latex{ 1 }; the probability of meeting is  \latex{\frac{11}{36}.}
The previous example shows that by illustrating continuously changing quantities in a coordinate system, the problem can be translated into a computation of geometric probability.

Notice that in regard of the problem only the order of the points is important; their specific place on the circle is not. Thus \latex{ 6 } points can be placed in \latex{ 5! } different orders.

The triangles \latex{ ABC } and \latex{ DEF } are disjoint if the points \latex{ A }, \latex{ B }, \latex{ C } can be separated by a line from points \latex{ D }, \latex{ E }, \latex{ F }. The order of the points on both sides of the line are indifferent, that is, the order of points \latex{ A }, \latex{ B }, \latex{ C } is indifferent as well as the order of points \latex{ D }, \latex{ E }, \latex{ F }, therefore there are \latex{3!\times 3!} possibilities.

Thus the probability in question is

\latex{\frac{3!\times 3!}{5!}=\frac{36}{120}=\frac{3}{10}.}

Upon first glance this looked like a geometric probability problem, but it turned out that it was a pure combinatorics problem.