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Triangle From a Ruler

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID:#26982
Fun:*** (2.46)
Difficulty:*** (2.79)
Submitted By:phrebh*us****
Corrected By:phrebh




Fred had a ruler that was exactly 12 inches long. His second cousin, Pop, was practicing with her Samurai sword and made two straight slashes at arbitrary spots on the ruler, cutting it into three pieces. What is the probability that the pieces of Fred's ruler can form a triangle?


To find the answer, we will use two random variables, a and b, on a common graph with rectangular axes. Both variables are uniformly distributed between 0 and 12, so we have a probability square which each side equal to 12. We will find regions on the square that lead to three lengths that can form a triangle.

Let us suppose that the two cuts are made at some lengths a and b from one end. If b > a, then the three lengths will be a, b-a, and 12-b. In order to create a triangle, any two of these lengths needs to be greater than the third. So,

a + b-a > 12-b or b > 6
b-a + 12-b > a or a < 6
a + 12-b > b-a or b < a+6

When we plot these on our graph, we get a triangle with points at (6,6), (0,6), and (6,12). The triangle formed has an area of 1/8th of the square.

If we then consider the case of a > b, then we get a similar triangular region reflected over the x=y diagonal of our graph. That triangle also has an area of 1/8th of the square.

Therefore the total area of our two regions is given by 1/8 + 1/8 = 1/4 and so the probability of Pop having created pieces capable of making a triangle is 25%.

The calculation of the probability that Fred lost a finger or two is left up to the reader.


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