document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1146498>Question 1146498</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Three cards are drawn at random without replacement.  Without knowing the values of the first two cards, what is the probability of the third card having value of 10 and/or being a face card? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #767798 by </B><A HREF=/tutors/aboutme.mpl?userid=VFBundy><B>VFBundy(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=VFBundy><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=VFBundy><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=767798\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> There are 16 cards that have a value of 10: the 10, J, Q, and K of all four suits.<BR>\n" );
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document.write( "Chances both of the first two cards have a \"10\" value = 16/52 * 15/51 = 20/221<BR>\n" );
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document.write( "Chances first card has a \"10\" value and second card does not have a \"10\" value = 16/52 * 36/51 = 48/221<BR>\n" );
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document.write( "Chances first card does not have a \"10\" value and second card has a \"10\" value = 36/52 * 16/51 = 48/221<BR>\n" );
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document.write( "Chances neither of the first two cards have a \"10\" value = 36/52 * 35/51 = 105/221<BR>\n" );
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document.write( "Continuing...<BR>\n" );
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document.write( "20/221 of the time, two \"10\" cards have been drawn, leaving 14 \"10\" cards remaining out of the 50 remaining cards.  Or, a 14/50 chance the third card will be a \"10\" card.<BR>\n" );
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document.write( "96/221 of the time, one \"10\" card has been drawn, leaving 15 \"10\" cards remaining out of the 50 remaining cards.  Or, a 15/50 chance the third card will be a \"10\" card.  <i>(The 96/221 number comes from adding together the chance where a \"10\" card is drawn first and the chance where a \"10\" card is drawn second.)</i><BR>\n" );
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document.write( "105/221 of the time, zero \"10\" cards have been drawn, leaving all 16 \"10\" cards remaining out of the 50 remaining cards.  Or, a 16/50 chance the third card will be a \"10\" card.<BR>\n" );
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document.write( "So...<BR>\n" );
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document.write( "20/221 of the time, there is a 14/50 probability the third card will be a \"10\" card.<BR>\n" );
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document.write( "96/221 of the time, there is a 15/50 probability the third card will be a \"10\" card.<BR>\n" );
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document.write( "105/221 of the time, there is a 16/50 probability the third card will be a \"10\" card.<BR>\n" );
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document.write( "The math is as follows:<BR>\n" );
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document.write( "(20/221) * (14/50) = 28/1105<BR>\n" );
document.write( "(96/221) * (15/50) = 144/1105<BR>\n" );
document.write( "(105/221) * (16/50) = 168/1105<BR>\n" );
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document.write( "Add together 28/1105 + 144/1105 + 168/1105 and you get 340/1105.  This fraction reduces to 4/13.<BR>\n" );
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document.write( "So, there is a 4/13 probability the third card will be a \"10\" card. </SPAN>\n" );
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