Category:Bertrand's paradox

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<nowiki>Paradoja de Bertrand; paradoxe de Bertrand; парадокс Бертрана; Bertrand-Paradoxon; Nghịch lý Bertrand (xác suất); Paradacsa Bertrand; پارادوکس برتراند; 伯特蘭悖論 (概率論); Bertrano paradoksas; Paradoxul lui Bertrand (teoria probabilităților); ベルトランの逆説; paradoks Bertranda; הפרדוקס של ברטראן; paradox van Bertrand; парадокс Бертрана; paradosso di Bertrand; 베르트랑의 역설; Bertrand paradox; مفارقة برتراند; Bertranda paradokso; Բերտրանի պարադոքս; problema dentro de la interpretación clásica de la teoría de la probabilidad; Paradoxon; 확률론에서 발생하는 역설 중의 하나; problem within the classical interpretation of probability theory: what is the probability that a randomly chosen chord on a circle is longer than a side of an equilateral triangle inscribed in the circle?; paradokso aŭ problemo de probabloteorio: kiom estas la probablo ke hazarde elektita ŝnuro sur iu cirklo estas pli longa ol latero de egallatera triangulo enskribita en tiu cirklo?; problem within the classical interpretation of probability theory: what is the probability that a randomly chosen chord on a circle is longer than a side of an equilateral triangle inscribed in the circle?; kansrekening; парадокс Бертрана (вероятность); ベルトランのパラドックス; ベルトランの逆理; paradoxe de bertrand; Бертрана парадокс (Теорія ймовірностей); парадокс Бертрана (теорія ймовірностей); paradokso de Bertrand; הפרדוקס של ברטרנד; הפרדוקס של ברטרן; Bertrano uždavinys; Bertrand paradoksas; Bertrand uždavinys</nowiki>
paradosso di Bertrand 
problem within the classical interpretation of probability theory: what is the probability that a randomly chosen chord on a circle is longer than a side of an equilateral triangle inscribed in the circle?
Random Endpoints method: Pick 2 random points on the circumference of the circle; draw the chord joining them. To find the probability: rotate the triangle so its vertex coincides with one of the chord endpoints. The chord is longer iff the other chord endpoint lies on the arc between the endpoints of the triangle side opposite the 1st point. So the probability is ⅓.
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  • 1889
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File nella categoria "Bertrand's paradox"

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