document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1045969>Question 1045969</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If 5 cards are drawn from an ordinary deck of playing cards what is the probability that the cards will contain a pair or better? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #661483 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=661483\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Assumptions:<BR>\n" );
document.write( "1) Assuming a 52 card deck. A deck has 13 ranks of each of the four suits: clubs, diamonds, hearts, and spades.<BR>\n" );
document.write( "2) Assuming a hand of 5 cards, like your problem says.<BR>\n" );
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document.write( "Objective: Draw one pair (two of one face value, and 3 cards of different face <BR>\n" );
document.write( "values, not matching the pair):\r<BR>\n" );
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document.write( "                13 x 4C2 x 12C3 x 4^3<BR>\n" );
document.write( " Prob(1 pair) = ----------------------  =  0.422569 or 42.257%<BR>\n" );
document.write( "                        52C5  <BR>\n" );
document.write( "The denominator:<BR>\n" );
document.write( "is the number of ways of selecting 5 cards from 52:<BR>\n" );
document.write( "52C5 means 52 Choose 5. <BR>\n" );
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document.write( "The numerator:<BR>\n" );
document.write( "13 is the number of ways of choosing the face value of the pair, e.g. two 10's or two queens or whatever.  \r<BR>\n" );
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document.write( "4C2 is the number of ways of choosing the two cards from the four same-value cards the pack.\r<BR>\n" );
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document.write( "12C3 is the number of ways of choosing 3 different face values from the 12 still available - we have used up one face value for the pair. For argument's sake, suppose the 3 remaining face values are 3, 7, K\r<BR>\n" );
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document.write( "The 4^3 term arises in the following way. Suppose the face values are to be 3, 7, K. We have 4 ways of choosing the 3,since there are four 3's in the pack. Similarly, there are 4 ways of choosing the 7 and 4 ways of choosing the K, giving 4 x 4 x 4 = 4^3 ways of selecting the other 3 cards with a given set of face values.\r<BR>\n" );
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document.write( "Multiplying all these factors together gives the total number of ways <BR>\n" );
document.write( "that we could choose exactly one pair.  And this divided by 52C5 gives <BR>\n" );
document.write( "the required probability. </SPAN>\n" );
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