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A-cord-is-drawn-at-random-in-a-given-circle-Whats-the-probability-p-that-the-cord-is-longer-than-the-side-of-an-inscribed-equilateral-triangle-in-the-circle-




Question Number 60000 by malwaan last updated on 16/May/19
A cord is drawn at random  in a given circle .  Whats the probability p that  the cord is longer than the side  of an inscribed equilateral   triangle in the circle?
Acordisdrawnatrandominagivencircle.Whatstheprobabilitypthatthecordislongerthanthesideofaninscribedequilateraltriangleinthecircle?
Commented by mr W last updated on 16/May/19
(1/3)
13
Commented by MJS last updated on 17/May/19
this is called Bertrand paradoxon. dependent  on the chosen method of randomization  we get p=1/2, 1/3 or 1/4. look it up!
thisiscalledBertrandparadoxon.dependentonthechosenmethodofrandomizationwegetp=1/2,1/3or1/4.lookitup!
Commented by malwaan last updated on 17/May/19
but sir  what is the best one?
butsirwhatisthebestone?
Commented by mr W last updated on 17/May/19
https://en.m.wikipedia.org/wiki/Bertrand_paradox_(probability)
Commented by MJS last updated on 17/May/19
there′s no best one. we cannot give an  answer. it′s undefined.
theresnobestone.wecannotgiveananswer.itsundefined.
Commented by mr W last updated on 17/May/19
I think the question is defined if it is:  Two points A and B are choosen at  random on a given circle.  What is the probability that the cord  AB is longer than l with 0<l<diameter.
Ithinkthequestionisdefinedifitis:TwopointsAandBarechoosenatrandomonagivencircle.WhatistheprobabilitythatthecordABislongerthanlwith0<l<diameter.
Commented by MJS last updated on 17/May/19
yes. you have to define the method
yes.youhavetodefinethemethod
Commented by ajfour last updated on 17/May/19
Commented by malwaan last updated on 18/May/19
whats the answer to your  method sir mr W ?
whatstheanswertoyourmethodsirmrW?
Commented by mr W last updated on 18/May/19
the positions of A and B can be expressed  as θ_A  and θ_B . ∣θ_A −θ_B ∣ is equally   distributed over 0°−180°. the probability  for  ∣θ_A −θ_B ∣≥120° is therefore  ((180−120)/(180))=(1/3).
thepositionsofAandBcanbeexpressedasθAandθB.θAθBisequallydistributedover0°180°.theprobabilityforθAθB∣⩾120°istherefore180120180=13.
Commented by malwaan last updated on 19/May/19
thank you sir
thankyousir

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