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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?

$${A}\:{cord}\:{is}\:{drawn}\:{at}\:{random} \\ $$$${in}\:{a}\:{given}\:{circle}\:. \\ $$$${Whats}\:{the}\:{probability}\:\boldsymbol{{p}}\:{that} \\ $$$${the}\:{cord}\:{is}\:{longer}\:{than}\:{the}\:{side} \\ $$$${of}\:{an}\:{inscribed}\:{equilateral}\: \\ $$$${triangle}\:{in}\:{the}\:{circle}? \\ $$

Commented by mr W last updated on 16/May/19

(1/3)

$$\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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!

$$\mathrm{this}\:\mathrm{is}\:\mathrm{called}\:\mathrm{Bertrand}\:\mathrm{paradoxon}.\:\mathrm{dependent} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{chosen}\:\mathrm{method}\:\mathrm{of}\:\mathrm{randomization} \\ $$$$\mathrm{we}\:\mathrm{get}\:{p}=\mathrm{1}/\mathrm{2},\:\mathrm{1}/\mathrm{3}\:\mathrm{or}\:\mathrm{1}/\mathrm{4}.\:\mathrm{look}\:\mathrm{it}\:\mathrm{up}! \\ $$

Commented by malwaan last updated on 17/May/19

but sir what is the best one?

$${but}\:{sir} \\ $$$${what}\:{is}\:{the}\:{best}\:{one}? \\ $$

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.

$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{best}\:\mathrm{one}.\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{give}\:\mathrm{an} \\ $$$$\mathrm{answer}.\:\mathrm{it}'\mathrm{s}\:\mathrm{undefined}. \\ $$

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.

$${I}\:{think}\:{the}\:{question}\:{is}\:{defined}\:{if}\:{it}\:{is}: \\ $$$${Two}\:{points}\:{A}\:{and}\:{B}\:{are}\:{choosen}\:{at} \\ $$$${random}\:{on}\:{a}\:{given}\:{circle}. \\ $$$${What}\:{is}\:{the}\:{probability}\:{that}\:{the}\:{cord} \\ $$$${AB}\:{is}\:{longer}\:{than}\:{l}\:{with}\:\mathrm{0}<{l}<{diameter}. \\ $$

Commented by MJS last updated on 17/May/19

yes. you have to define the method

$$\mathrm{yes}.\:\mathrm{you}\:\mathrm{have}\:\mathrm{to}\:\mathrm{define}\:\mathrm{the}\:\mathrm{method} \\ $$

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 ?

$${whats}\:{the}\:{answer}\:{to}\:{your} \\ $$$${method}\:{sir}\:{mr}\:{W}\:? \\ $$

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).

$${the}\:{positions}\:{of}\:{A}\:{and}\:{B}\:{can}\:{be}\:{expressed} \\ $$$${as}\:\theta_{{A}} \:{and}\:\theta_{{B}} .\:\mid\theta_{{A}} −\theta_{{B}} \mid\:{is}\:{equally}\: \\ $$$${distributed}\:{over}\:\mathrm{0}°−\mathrm{180}°.\:{the}\:{probability} \\ $$$${for}\:\:\mid\theta_{{A}} −\theta_{{B}} \mid\geqslant\mathrm{120}°\:{is}\:{therefore} \\ $$$$\frac{\mathrm{180}−\mathrm{120}}{\mathrm{180}}=\frac{\mathrm{1}}{\mathrm{3}}. \\ $$

Commented by malwaan last updated on 19/May/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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