document.write( "Question 681564: In English:
\n" ); document.write( "Eight cards are extracted without replacement from an ordinary deck. find the probability of getting exactly three ACEs or exactly three kings (or both).\r
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\n" ); document.write( "\n" ); document.write( "ocho cartas se extraen sin reposición de una baraja ordinaria. encontrar la probabilidad de obtener exactamente tres ases o exactamente tres reyes (o ambos). \n" ); document.write( "
Algebra.Com's Answer #422705 by Edwin McCravy(20067)\"\" \"About 
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document.write( "En español:\r\n" );
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\n" ); document.write( "Ocho tarjetas se extraen sin reemplazo de una baraja ordinaria. hallar la probabilidad de obtener exactamente tres ases o Reyes exactamente (o ambos).
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document.write( "Hay 4 ases en una baraja normal.\r\n" );
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document.write( "Hay maneras de C(4,3) 3 Ases\r\n" );
document.write( "Hay maneras de 5 tarjetas que no sean ases C(48,5)\r\n" );
document.write( "Hay formas C(52,8) para obtener las 8 tarjetas\r\n" );
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document.write( "P(EXACTAMENTE 3 ASES) = \"%28%22C%284%2C3%29%22%2A%22C%2848%2C5%29%22%29%2F%22C%2852%2C8%29%22\" =.0091014867\r\n" );
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document.write( "Hay 4 Reyes, por lo que la probabilidad de obtener exactamente 3 Reyes es el mismo:\r\n" );
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document.write( "P(EXACTAMENTE 3 REYES) =.0091014867\r\n" );
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document.write( "Pero también tenemos el número de as 3 y 3 Reyes,\r\n" );
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document.write( "Hay maneras de C(4,3) 3 Ases\r\n" );
document.write( "Hay maneras de C(4,3) 3 Reyes\r\n" );
document.write( "Hay maneras de C(44,2) 2 tarjetas que no sean ases ni Reyes\r\n" );
document.write( "Hay formas C(52,8) para obtener las 8 tarjetas\r\n" );
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document.write( "P (exactamente 3 Ases y 3 Reyes) = \"%28%22C%284%2C3%29%22%2A%22C%284%2C3%29%22%2AC%2844%2C2%29%29%2F%22C%2852%2C8%29%22\" =\r\n" );
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document.write( ".00000002126138057\r\n" );
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document.write( "Ahora utilizamos la ecuación:\r\n" );
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document.write( "P (EXACTAMENTE 3 ASES O REYES EXACTAMENTE 3 (O AMBOS)) =\r\n" );
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document.write( "P(EXACTAMENTE 3 ASES) + P (EXACTAMENTE 3 REYES) - P (EXACTAMENTE 3 ASES Y REYES EXACTAMENTE 3)\r\n" );
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document.write( "=.0091014867 +.0091014867 -.00000002126138057 =.0182029522\r\n" );
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document.write( "In Engish:\r\n" );
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\n" ); document.write( "Eight cards are extracted without replacement from an ordinary deck. find the probability of getting exactly three ACEs or exactly three kings (or both).
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document.write( "There are 4 ACES in an ordinary deck of cards.\r\n" );
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document.write( "There are C(4,3) ways to get 3 ACES\r\n" );
document.write( "There are C(48,5) ways to get 5 cards which are not ACES\r\n" );
document.write( "There are C(52,8) ways to get any 8 cards\r\n" );
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document.write( "P(exactly 3 ACES) = \"%28%22C%284%2C3%29%22%2A%22C%2848%2C5%29%22%29%2F%22C%2852%2C8%29%22\" = .0091014867\r\n" );
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document.write( "There are 4 KINGS, so the probability of getting exactly 3 KINGS is the same:\r\n" );
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document.write( "P(exactly 3 KINGS) = .0091014867\r\n" );
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document.write( "But we must also get the number of 3 ACE and 3 KINGS,\r\n" );
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document.write( "There are C(4,3) ways to get 3 ACES\r\n" );
document.write( "There are C(4,3) ways to get 3 KINGS\r\n" );
document.write( "There are C(44,2) ways to get 2 cards which are neither ACES nor KINGS\r\n" );
document.write( "There are C(52,8) ways to get any 8 cards\r\n" );
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document.write( "P(exactly 3 ACES and 3 KINGS) = \"%28%22C%284%2C3%29%22%2A%22C%284%2C3%29%22%2AC%2844%2C2%29%29%2F%22C%2852%2C8%29%22\" =  \r\n" );
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document.write( ".00000002126138057\r\n" );
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document.write( "Now we use the equation:\r\n" );
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document.write( "P(EXACTLY 3 ACES OR EXACTLY 3 KINGS (OR BOTH)) = \r\n" );
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document.write( "P(EXACTLY 3 ACES) + P(EXACTLY 3 KINGS) - P(EXACTLY 3 ACES AND EXACTLY 3 KINGS)\r\n" );
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document.write( "= .0091014867 + .0091014867 - .00000002126138057 = .0182029522\r\n" );
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document.write( "Edwin
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