document.write( "Question 971466: A bag contains 9 red marbles and 5 yellow marbles. You are asked to draw 3 marbles from the bag without replacement. In how many ways can you draw 2 reds and 1 yellow?\r
\n" ); document.write( "\n" ); document.write( "I do not know how to solve this problem. Please provide me with an explanation to the solution and the answer.
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Algebra.Com's Answer #593968 by Boreal(15235) $\"\"$ $\"About$

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prob 2 reds and 1 yellow Without replacement,the denominator changes, and so does the numerator if a given marble is chosen. There are 9 ways out of 14 to get a red first. Once a red has been chosen, there are 13 marbles left, but only 8 red ones.\r
\n" ); document.write( "\n" ); document.write( "(9/14)*(8/13)*(5/12)\r
\n" ); document.write( "\n" ); document.write( "There are 3 different ways this will occur, if the yellow is drawn 1st, 2nd, or third.\r
\n" ); document.write( "\n" ); document.write( "(5/14)(9/13)(8/12) is another way, but this is the same fraction when multiplied.
\n" ); document.write( "In all cases, however, the denominator is the same and the numerator will multiply out the same, even if it is over a different denominator.\r
\n" ); document.write( "\n" ); document.write( "(360/2184)*3=1080/2184 \r
\n" ); document.write( "\n" ); document.write( "P= 0.495 \n" ); document.write( "

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