document.write( "Question 326836: The probability that a person will have 0,1, or 2 dental checkups per year is 0.3, 0.6, and 0.1 repectively. If seven people are picked at random, what is the probability that two will have no checkups, four will have one checkup, and one will have two checkups in the next year?
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document.write( "A) 0.012
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document.write( "B) 0.018
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document.write( "C) 0.122
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document.write( "D) 0.588\r
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document.write( "All I can figure out and I am not sure if this is right is: (7!/2!*4!*1!) \n" );
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Algebra.Com's Answer #234048 by galactus(183) You can put this solution on YOUR website! This is a multinomial probability.\r \n" ); document.write( "\n" ); document.write( "You are partly correct.\r \n" ); document.write( "\n" ); document.write( "You just have to multiply by the number of respective probabilities.\r \n" ); document.write( "\n" ); document.write( " |