document.write( "Question 1030146: Hi,
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document.write( "Can someone please help me with this.\r
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document.write( "The chances of James winning a game are 0.55. What is the probability of James winning the game in 3rd trial?
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document.write( " 0.11
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document.write( " 0.42
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document.write( " 0.16
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document.write( " 0.33
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document.write( "I have got the answer as 0.16 but not confident. I have used BINOM.DIST(3,3,0.55,FALSE) in excel. \n" );
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Algebra.Com's Answer #645045 by Theo(13342)![]() ![]() You can put this solution on YOUR website! probability of winning is .55 \n" ); document.write( "probability of losing is 1 - .55 = .45 \n" ); document.write( "probability of winning in the third trial would be .45 * .45 * .55 = .111375 which rounds to .11.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "your are using the binomial distribution. \n" ); document.write( "that tells you how many successes you get out of 3 tries. \n" ); document.write( "probabillity of success = .55 \n" ); document.write( "probability of failure = .45\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that formula is p(x) = c(n,x) * p^x * q^(n-x)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "what you entered should get you c(3,3) * .55^3 * .45^0 = .166375\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "what that is telling you is the probably of winning all 3 games out of 3.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that's not the same as winning only the third game.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i think your solution is .11.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "it's the probability of losing the first 2 games times the probability of winning the third game.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |