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: Probability on triangle!  ( 7126 )
« : June 26, 2007, 08:34:50 PM From Poonam»



Probability on triangle!


If a Pole is broken into three pieces, what is the probability that the pieces can be put together to make a triangle?

 
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« #1 : July 02, 2007, 12:22:52 PM From preep»

1/3 ?? im not sure..
 ???but i need answr for this...
« #2 : July 02, 2007, 12:39:09 PM From PSM»

the method is tht....if sum of any two sides i greater than the other then only they can make a triangle.......but how is it possible to find the probability tht the broken parts of the pole satisfies the condition...
« #3 : August 17, 2007, 11:59:33 PM From Shashi»

If it satisfies the condition then the probability is one
« #4 : August 30, 2007, 08:38:00 PM From jigar_er_civil»

obviously 1 because 3 piece make triangle , but condition follow
« #5 : September 12, 2007, 10:46:01 AM From Rajesh»

The Probability will be 1/2
P(E) = 0.5

if we divide st line into 2n points and for only those cases when we will choose both the points from left side or right side of mid point and if one of the point is mid point the triangle could not be formed and all other cases are favorable cases. That means one point is from left side of mid point and other is from right side these are only cases when a triangle could be formed (we can divide a st line into three parts by only choosing two points on it).....so

and whenever sum of two sides will be equal to third one that means one point is the mid point of st line we cant draw the triangle but these r very less cases (2n cases) as compared to the favourable cases (n^2 cases) since n is tending to infinity so probability remain same. That is it will be half.
 

P(E) = 1- P(not forming any triangle)
      =  lim n-> infinity [1-  (nc2+nc2+2n)/(2n)c2]
      =  1-1/2
      =  1/2

Alternate soln.....

P(E) = favourable cases/total cases
      =  im n-> infinity [(nc1*nc1)/(2n)c(2)]
       =  im n-> infinity [ n^2/n(2n-1)]
       = im n-> infinity [n/(2n-1)]
        =  1/2


 
 
« #6 : September 12, 2007, 02:51:01 PM From vishwakarma.deepak»

Im with rajesh...


dividing line in three parts to make trangle, the probability should be 1/2...

guru pls confirm answer...

Deepak...
« #7 : September 12, 2007, 03:40:53 PM From Poonam»

Guys, I don't know the answer. Rajesh's proof seems convincing. Any contradictory argument?
« #8 : September 12, 2007, 11:27:25 PM From spazinvader»

I have a little doubt.
U have given that lim n-> infinity [1-  (nc2+nc2)/2nc2] right?
How come (nc2+nc2)/2nc2 becomes 1/2?
It should be 1 right?
If it is 1 then the probability will be 0.

Please would u tell me Rajesh?
« #9 : September 12, 2007, 11:58:36 PM From Rajesh»

Hi Spazinvader......I got your doubt
here nc2 + nc2 is not equal to 2nc2....in fact its my printing mistake
2nc2 is (2n)c(2)....ie (fact 2n)/(fact 2)*[fact (2n-2)] and nc2 is (n)c(2)

given .....ncr = fact n/fact r * fact (n-r) ....fact= factorial

so [(n)c(2) + (n)c(2)]/[(2n)c(2)] = n(n-1)/n(2n-1)
                             =  lim n-> infinity [(n-1)/(2n-1)]
                              =lim n-> infinity [(1-1/n)/(2-1/n)]
                              =  1/2

« #10 : September 14, 2007, 10:18:52 PM From spazinvader»

I got it Rajesh.Thanks for the info.
I have taken it as (nc2+nc2)/2(nc2).
Thanks for the full info.
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