document.write( "Question 1075656: A box contains six tickets, with letters: A, B, C, D, D, E. Note that there are 2 D’s in the box. Nine draws will be made at random with replacement. Find the probability that exactly four of the tickets drawn will be a D. Express your answer as a percent.\r
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document.write( "9 C 4 (2/6)^4 (4/6)^5 I am right so far... nobody replied yesterday... \n" );
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Algebra.Com's Answer #690469 by Theo(13342) You can put this solution on YOUR website! p(d) = 2/6 = 1/3\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "binomial probability is used with p = 1/3 and q = 1 - p = 2/3\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "binomial probability formula is p(x) = p^x * q^(n-x) * c(n,x)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "this becomes p(4) = (1/3)^4 * (2/3)^5 * c(9,4)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that looks like what you have.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "when you evaluate it, you get p(4) = .2048468221.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "from what i know, i believe you are correct.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |