Probability of a TriangleDate: 12/10/97 at 06:29:33 From: Chez Kwiatkowski Subject: Probability If I cut a 12-inch ruler with two arbitrary cuts, what is the probability that from the three pieces created I will be able to create a triangle? I came up with 25 percent probability. Is that correct? Chez Date: 12/10/97 at 08:32:13 From: Doctor Anthony Subject: Re: Probability These problems are best done by considering two random variables x and y on the usual rectangular axes, with both x and y uniformly distributed between 0 and 12. The probability space is a square of side 12, and we need to consider regions of the square that lead to three lengths which can form a triangle. Suppose that the breaks are made at points distance x and y from the one end. Then if y > x (we can consider the alternative later) the three lengths will be x, y-x and 12-y. For a triangle any two of these must be greater than the third, so x + y-x > 12-y y-x + 12-y > x x + 12-y > y-x y > 6 x < 6 y < x+6 If you plot these three boundaries you get a triangle with vertices at (6,6) (0,6) (6,12) and area equal 1/8 of the square. This assumed y > x. If instead x > y then we get a second region symmetrically placed below the diagonal y=x of the square and again having an area 1/8 the area of the square. So the total area of the square giving values of x and y which allow for a triangle to be formed is 1/8 + 1/8 = 1/4 . So the probability that a triangle can be formed is 1/4. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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