document.write( "Question 852124: If the probability of a tennis player serving an ace is 1/6, what's the probability that the player will serve 3 aces out of 8 serves?\r
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document.write( "I know it's independent probability. I did it like this:
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document.write( "for \"non-ace\" results, the prob. is 5/6
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document.write( "for 8 serves, the prob of 3 aces is:\r
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document.write( "1/6 * 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 3125/1,679,616 = approx. 0.00186, or 0.186%, or 186/1000.\r
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document.write( "Am I even close? \n" );
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Algebra.Com's Answer #513162 by Fombitz(32388)![]() ![]() You can put this solution on YOUR website! Yes and no. \n" ); document.write( "I think you've got the concepts but you're missing a lot of terms. \n" ); document.write( "Since you don't know when the ace is served, the example you gave is only one possible outcome. \n" ); document.write( "It could have been \n" ); document.write( "AAANNNNN \n" ); document.write( "or \n" ); document.write( "ANAANNNN \n" ); document.write( "or \n" ); document.write( "NNNNNAAA \n" ); document.write( "So you need to capture all of the possible outcomes that each individually have a probability of \n" ); document.write( "You can use Pascal's triangle with 9 terms (1,8,28,56,70,56,28,8,1). \n" ); document.write( "So for no aces, there is 1 term (NNNNNNNN) \n" ); document.write( "For 1 ace, 8 terms (ANNNNNNN, NANNNNNN, .... , NNNNNNNA) \n" ); document.write( "For 2 aces, 28 terms (too many terms to write out) \n" ); document.write( "For 3 aces, 56 terms. \n" ); document.write( "So then,\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "The other terms are, \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If you add all the terms, they do sum to 1. \n" ); document.write( " |