La paradoja de Bertrand

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<span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>Formulada por primera vez por el matem&aacute;tico franc&eacute;s&nbsp;</span><a href=»http://www-history.mcs.st-and.ac.uk/Biographies/Bertrand.html» style=»box-sizing: border-box; color: rgb(52, 152, 219) !important; text-decoration: none; background: 0px 0px;»>Joseph Louis Fran&ccedil;ois Bertrand</a><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>en su&nbsp;</span><i style=»box-sizing: border-box;»>Calcul des probabilit&eacute;s</i>&nbsp;(1888)<span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>, la &lsquo;paradoja de Bertrand&rsquo; es una paradoja de la teor&iacute;a de la probabilidad que se enuncia del siguiente modo:</span>
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<span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>Consideremos una circunferencia&nbsp;</span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»><i style=»box-sizing: border-box;»><span style=»box-sizing: border-box; font-weight: 700;»>C</span></i></span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>&nbsp;y una cuerda &ndash;un segmento cuyos extremos est&aacute;n situados sobre&nbsp;</span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»><span style=»box-sizing: border-box; font-weight: 700;»>C</span></span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>&ndash;&nbsp;</span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»><i style=»box-sizing: border-box;»><span style=»box-sizing: border-box; font-weight: 700;»>AB</span></i></span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>&nbsp;trazada al azar sobre ella.</span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>&iquest;Cu&aacute;l es la probabilidad&nbsp;</span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»><i style=»box-sizing: border-box;»><span style=»box-sizing: border-box; font-weight: 700;»>p</span></i></span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>&nbsp;de que esta cuerda sea m&aacute;s</span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>larga que el lado del tri&aacute;ngulo equil&aacute;tero inscrito en la circunferencia?</span>
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<span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>Abordando el problema de diferentes maneras, Bertrand obtuvo tres resultados distintos. &iquest;C&oacute;mo es posible? La pregunta clave es:&nbsp;</span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»><i style=»box-sizing: border-box;»>&iquest;</i></span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>Qu&eacute; significa</span><span style=»box-sizing: border-box; color: rgb(0, 0, 0);»><i style=»box-sizing: border-box;»>&nbsp;trazar una cuerda al azar?</i></span>
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<span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>Veamos esas tres respuestas.</span>
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<span style=»box-sizing: border-box; color: rgb(0, 0, 0);»>Para seguir leyendo ir al siguiente enlace: <a href=»http://culturacientifica.com/2013/10/30/la-paradoja-de-bertrand/» target=»_blank»>Cuaderno de Cultura Cient&iacute;fica</a></span>
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