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Re: probability of a triangle (SPOILER)
Posted:
Jul 12, 2004 7:49 PM
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I believe that the answer to part (b) should still be 1/4.
1. Heuristically or intuitively, although the 2 sequences of events
as "described" in part (a) and part (b) may "appear" to be different,
in the end they really amount both to the same event of breaking a
unit length into 3 pieces, whether by making 2 "simultaneous" cuts
(part a) or making 1 cut followed by a second one (part b).
2. Mathematically, suppose the first cut in (b) results in 2 pieces,
and the piece picked to be further broken has length X, with the other
piece of length c = (1-X). Then supposed X is further divided in 2
pieces of length b = y and a = (X-y).
Clearly, 0 < y < X
In the cartesian plane (X, y), the domain of all possible points
(X, y) is the right triangle intersection of the 3 areas y>0, y<X and
X<1.
The subdomain of all points (X, y) resulting in a, b and c forming
a triangle must satisfy the 3 conditions a < 1/2, b < 1/2 and c < 1/2,
or:
y > X - 1/2, y < 1/2 and X > 1/2
The area of this triangular subdomain is 1/4 the area of the entire
domain, thus the probability remains 1/4, conforming with what
intuition indicated.
On Jun 7 12:48:12 1996, Daniel A. Asimov wrote:
>In article <9604298333.AA833373730@ccmail.odedodea.edu>
Pat_Ballew@ccmail.odedodea.edu writes:
>> b) If the length is broken at a random point, and then one of
the two
>> pieces is randomly selected and broken at a random point on its
length
>> what is the probability that the three pieces will form a
triangle
>>
>> Pat Ballew
>> Misawa, Japan
>> Pat_Ballew@ccmail.odedodea.edu
>-----------------------------------------------------------------------
>
>Here's my answer to question b):
>
>After the first random break, we can (without loss of generality)
skip the step
>of randomly selecting one of the pieces, since the distribution of
the left
>and right pieces will be identical anyway. This simplifies the
problem a bit.
>
>Say we always choose the leftmost piece; call its length x.
>
>At this stage we have 2 pieces: [0,x] and [x,1].
>
>Now we want to break [0,x] "at a random point", which amounts to
choosing
>(independent of x) a number y at random in the interval [0,1], where
y
>represents the fraction of x where the break will occur.
>
>So, we end up with 3 pieces: [0,xy], [xy, x], and [x,1]. Their
lengths are
>clearly xy, x-xy = x(1-y), and 1-x.
>
>The distribution on x and y is uniform on the unit square
[0,1]x[0,1].
>
>And the region where the 3 pieces form a triangle corresponds (as
Mark Burkey,
>and via e-mail, Pat Ballew have pointed out) to precisely the
condition that
>all 3 sides are of length < 1/2.
>
>The subset of the square, then, where xy < 1/2, x(1-y) < 1/2, and 1-x
< 1/2
>(i.e. x > 1/2) is the thorn-shaped region bounded on the left by the
line
>x = 1/2 and on the right by the 2 hyperbolas xy = 1/2 and x(1-y) =
1/2.
>
>Calculus then gives the area of this region -- which must be the
probability we
>are seeking -- as Prob(triangle) = ln(2) - 1/2 =
.1931471805599453094172321....
>
>--Dan Asimov
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