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As we have seen in Problem 2, we saw that calculating the probability of a given event depends on the definition of "random". Consider the following scenario:
Problem
Kate and Susan each have stick and want to calculate the probability of forming a triangle if the stick is broken in two places. Kate independantly chooses two random points on the stick. Susan takes a different approach. She breaks the original stick once and then breaks the longer of the two halves again. Thus the second point depends on the first. How much more likely is Susan able to form a triangle than Kate?
Solution
Let's assume that the twig has length 1 and let B1 and let B2 be the points where the stick is broken.
To successfully form a triangle, the sum of the lengths of the two sides must be greater than the length of the hypotenuse. Suppose b1 < b2. Then the three sides of the triangle have lengths b2, (b1 - b2), and (1 - b1).
These two conditions imply:
(1) b1 > 1/2
(2) b1 - b2 < 1/2
(3) b2 < 1/2
Since Kate chose her each point at random from the interval (0,1), the sample space for her experiment is the unit square. Susan's sample space, however, is restricted since her second point is chosed in relation to the first. Can you see why her sample space is restricted to only the yellow space in Figure 2? Using these equations, the event space is labeled in green below.
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Kate |
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Susan |
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Figure 1 |
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Figure 2 |
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By a ratio of areas, the probability that Kate will form a triangle is 1/4 and the probability that Susan will form a triangle is 1/3. Thus Susan is 4/3 times as likey to succeed.
There is a snag in our reasoning. The problem in Susan's case is that equal areas are not equally probable. To illustrate this, let's first assume that they are. Calclulate the area of vertical strips of width 
that are located on the x-axis at two points 0 < a <b < 1/2. In Susan's case the probability that 
is greater than the probability that 
. Can you see why? This violates our assumption that the points in the event space are uniformly distributed and thus the equal areas are not equally probable.
In order to correct this, we must attribute a larger probability to the shorter interval. Supposing that 
, 
is then uniformly distributed on the interval 
rather than (0,1). Thus the probability that 
.
It follows that the probability that Susan successfully builds a triangle is given by:
Notice this value is not equal to our original answer of 1/3.
The lack of uniform distribution leads us to the study of density. Because the points are not evenly distributed throughout the sample space, they may be more dense in some areas than others. The figure below is a representation of the inequality of probabilities that lead to a density function in Susan's experiment.
