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geometric probability examples geometric probability examples random events that take place in continuous sample space may invoke geometric imagery for at least two reasons due to the nature of the problem or due the the nature of the solution some problems like buffon s needle birds on a wire bertrand s paradox or the problem of the stick broken into three pieces do by their nature arise in a geometric setting the latter also admits multiple reformulations which require comparison of the areas of geometric figures in general we may think of geometric probabilities as non-negative quantities not exceeding 1 being assigned to subregions of a given domain subject to certain rules if function is an expression of this assignment defined on a domain d then for example we require 0 a 1 a d and d 1 the function is usually not defined for all a d those subsets of d for which is defined are the random events that form a particular sample spaces very often is defined by means of the ratio of areas so that if a is defined as the area of set a then one may set a a d know more about d2y/dx2 tutorcircle.com page no 1/4

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p. 2

problem 1 two friends who take metro to their jobs from the same station arrive to the station uniformly randomly between 7 and 7:20 in the morning they are willing to wait for one another for 5 minutes after which they take a train wether together or alone what is the probability of their meeting at the station in a cartesian system of coordinates s t a square of side 20 minutes represents all the possibilities of the morning arrivals of the two friends at the metro station the gray area a is bounded by two straight lines t s 5 and t s 5 so that inside a |s t 5 it follows that the two friends will meet only provided their arrivals s and t fall into region a the probability of this happening is given by the ratio of the area of a to the are of the square [400 15×15/2 15×15/2 400 175/400 7/16 problem 2 three points a b c are placed at random on a circle of radius 1 what is the probability for abc to be acute fix point c the positions of points a and b are then defined by arcs and extending from c in two directions a priori we know that 0 2 the favorable for our problem values of and as subtending acute angles satisfy 0 and 0 their sum could not be less than as this would make angle c obtuse therefore the situation is presented in the following diagram where the square has the side 2 region d is the intersection of three half-planes 0 0 and 2 this is the big triangle in the above diagram the favorable events belong to the shaded triangle which is the intersection of the half-planes and the ratio of the areas of the two is obviously 1/4 read more about real number properties worksheets tutorcircle.com page no 2/4

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p. 3

now observe that unless the random triangle is acute it can be thought of as obtuse since the probability of two of the three points a b c forming a diameter is 0 for bc to be a diameter one should have which is a straight line with zero as the only possible assignment of area thus we can say that the probability of abc being obtuse is 3/4 for an obtuse triangle the circle can be divided into two halves with the triangle lying entirely in one of the halves it follows that 3/4 is the answer to the following question three points a b c are placed at random on a circle of radius 1 what is the probability that all three lie in a semicircle references d a klain g c rota introduction to geometric probability cambridge university press 1997 a a sveshnikov problems in probability theory mathematical statistics and theory of random functions dover 1978 a m yaglom i m yaglom challenging mathematical problems with elementary solutions dover 1987 tutorcircle.com page no 3/4 page no 2/3

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p. 4

thank you tutorcircle.com

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