Onmatu- General Math Learning Article

Geometric probabilities

        Let's think about of a segment of line, which has 10 cm length. And let's put a random point on this line - i.e. so that this point to have equal chances to fall in one place like in any other. What is the probability that this point to fall in the middle of this segment line? The number of points on this segment of line is infinite - or, more clearly, as the segment is divided into segments increasingly smaller, the number of these segments increases indefinitely. Here, we don't deal with a finite number of cases. It has little sense to seek the likelihood that the point will fall exactly in the center. And this is because we cannot decide if this event ever occurred or not. However, is there a logical sense to ask ourselves what is the probability that the point to fall in a segment of line of 0.01 cm length, placed in the center? It is clear that a reasonable answer would be 0.01/10.      
        A beautiful problem of geometric probabilities is the next one we propose. Let's consider a segment of line AB and a point placed somewhere at the right of the middle of our segment - that divides the segments in two parts of lengths "a" and "b", with "b" smaller than "a". Now let's choose a point C at random on the longest segment, and another point D, also at random, on the shortest segment. These two points, C and D, divide our initial segment of line in three parts - AC, CD, DB. What is the probability that with any of these three segments to build a triangle? This problem is not so simple, but the answer "p=b/(2a)"is, instead, very simple and elegant.
        Another classical problem of geometric probabilities is the problem of Bufon. We suppose that on a plane surface parallel lines are drawn at the distance "a" one of another (as cracks on a floor planks). We throw on this plane, at random, a bar (a needle or match needle) of a length "c" that is smaller than "a". What is the probability that the object thrown to intersect one of the lines? The answer is "2c/(πa)". Therefore, throwing the bar a large number of times N and noting the number of crossing with "m", can be calculated experimentally the value of π, with the formula "π=2cN/(am)". At first sight it seems almost magic that the answer involve the number π. However, do not forget that π is always related with the circles and the angles. If the center of the needle or the needle match falls in a random point, its extremities can fall anywhere on a circle which surrounds that point because from the conditions of our problems, it's equal probable to have an orientation as well as any other. This way the circle comes in the problem - and with him also π.
With more than 100 years ago a certain teacher named Wolf from Frankfurt he had the patience to throw a needle 5,000 times; the needle had 36 mm length and the plane was crossed by parallel lines at the distance of 45 mm one from each other. He observed that the needle had crossed the lines for 2,532 times; from here we get an approximation of the value of
π (with only 0.6% bigger than the exact value of 3.1416). At the end of the last century it was said that a certain captain Fox has made over 1,120 throws which gives a value for π of 3.1419, and later a mathematician named Lazzarini made 3,408 throws getting a value of 3.1415929, which differs from the real value π=3.14159265 at the seventh decimal. It's easy to guess that it was something strange in the experiment of Lazzarini. Thank you
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