document.write( "Question 478909: Dear Math teacher, \r
\n" ); document.write( "\n" ); document.write( "Could you help me please with the following problem? \r
\n" ); document.write( "\n" ); document.write( "Three cards are drawn from a pack of 52, each card being replaced before the next one is drawn. Compute the probability p that all are spades. A pack of 52 cards includes 13 spades. \r
\n" ); document.write( "\n" ); document.write( "Here is how I did it: \r
\n" ); document.write( "\n" ); document.write( "p = 13C3 / 52C3 = 286/22,100 = 0.0129 or 1.29%. However, this is NOT the right answer. My textbook's solution is solved like this: (13/52)^3 = 1/64 = 0.0156 or 1.56%. \r
\n" ); document.write( "\n" ); document.write( "Please let me know why my way does NOT give me the right answer. \r
\n" ); document.write( "\n" ); document.write( "Thank you so much as always. \r
\n" ); document.write( "\n" ); document.write( "Yours sincerely, \r
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\n" ); document.write( "\n" ); document.write( "Ivanka \n" ); document.write( "

Algebra.Com's Answer #328166 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

 
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document.write( "Hi,
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document.write( "Three cards are drawn from a pack of 52, 
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document.write( "\"each card being replaced before the next one is drawn\"
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document.write( "P(all spades) = P(1st one a spade 'and' 2nd one a spade 'and' 3rd one a spade)
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document.write( "              = 13/52*13/52*13/52 = 1/4*1/4*1/4 = 1/64 = .0156 OR 1.56%
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document.write( "\"Without replacememt!\"
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document.write( "P(all spades) = 13/52*12/51*11/50 = 1716/132,600 = 286/22,100 = .0129 OR 1.29%\r
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document.write( "What that says 'statistically' is that replacing the card...ups the chance of
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document.write( "all spades being drawn \n" );
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