Re: puzzle question
- From: quasi <quasi@xxxxxxxx>
- Date: Sun, 01 Jan 2006 14:18:10 -0500
On Sun, 01 Jan 2006 13:31:29 EST, "M.A.Fajjal" <h2maf@xxxxxxxxx>
wrote:
>> On Sun, 01 Jan 2006 02:02:08 -0500, quasi
>> <quasi@xxxxxxxx> wrote:
>>
>> >On 31 Dec 2005 22:46:54 -0800, "mathbee"
>> <buntygill@xxxxxxxxx> wrote:
>> >
>> >>Given a stick of length L, what is the chance that
>> you can break it
>> >>into 3 pieces that can form a triangle?
>> >>
>> >>My answer is 1/8 (many other people got the same
>> answer) but the
>> >>interviewer said it is 1/3.
>> >>
>> >>Any ideas?
>> >
>> >It depends on the assumptiuons for the how the
>> breaking is done.
>> >
>> >Assuming the 2 breakpoints are chosen independently
>> and with a uniform
>> >distribution, I don't get 1/8 or 1/3 -- I get a
>> probability of 1/4.
>> >
>> >quasi
>>
>> Ok, quick challenge problem ...
>>
>> With the same assumptions, what is the probability
>> that you can form
>> an acute triangle?
>>
>let the angls of triangle are a, b, c
>
>a+b+c=180
>
>assume the angles are integers
>
>Acute triangle cases = 44/179
>
>Probability of acute triangle for the same case assumptions = 1/4 * 44/179 = 11/179
That's not the correct answer.
The assumptions are that the break points of the stick are independent
and uniformly distributed. That doesn't imply uniform distribution on
the angles. In fact, it's clear that the angles are not even
independent.
quasi
.
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