document.write( "Question 342868: A file cabinet contains 3 blue and 6 red folders. If folders are randomly selected, \"without replacement\", what is the probability of selecting:
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document.write( "A. Two red folders?
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document.write( "B. Three blue folders?
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document.write( "C. The first folder blue and the second folder red?
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document.write( "D. A red and a blue folder?\r
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document.write( "I got:\r
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document.write( "A: 2/9= 22.22%
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document.write( "B: 3/7= 42.85%
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document.write( "C: 0 (because the blue folders were taken out with replacement?)
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document.write( "D: 0 (because again, there are no blue folders left?) \n" );
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Algebra.Com's Answer #245374 by edjones(8007)![]() ![]() You can put this solution on YOUR website! There is replacement between questions! \n" ); document.write( ". \n" ); document.write( "A: 6/9 * 5/8 = 5/12 or .4167 \n" ); document.write( ". \n" ); document.write( "B: 3/9 * 2/8 * 1/7 = 1/84 \n" ); document.write( "OR \n" ); document.write( "nCr=combination or n things taken r at a time.= n!/((n-r)!r!) (easier with larger numbers.) \n" ); document.write( "3C3/9C3 = 1/84 \n" ); document.write( ". \n" ); document.write( "C: 3/9 * 6/8 = 1/4 \n" ); document.write( ". \n" ); document.write( "D: (6C1 + 3C1)/9C2 \n" ); document.write( "(6+3)/36 = 1/4 \n" ); document.write( ". \n" ); document.write( "Ed \n" ); document.write( " |