document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1050727>Question 1050727</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Having trouble with this homework! \r<BR>\n" );
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document.write( "Compute the probability of randomly drawing five cards from a deck and getting 3 Aces and 2 Kings. (Give your answer as a fraction.) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #666342 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=666342\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you draw 5 cards from the deck.\r<BR>\n" );
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document.write( "the probability that you would get 3 aces and 2 kings would be calculated as follows:\r<BR>\n" );
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document.write( "the number of ways you can get 3 aces out of 4 aces is c(4,3) = 4.\r<BR>\n" );
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document.write( "the number of ways you can get 2 kings out of 4 kings is c(4,2) = 6.\r<BR>\n" );
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document.write( "the number of ways you can get 2 kings and 3 aces is 4 * 6 = 24.\r<BR>\n" );
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document.write( "the number of ways you can get 5 cards out of a deck of 52 cards is c(52,5) = 2598960.\r<BR>\n" );
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document.write( "the probability of getting 3 aces and 2 kings when you draw 5 cards from the deck is 24 / 2598960 = 9.234463016 * 10^-6.\r<BR>\n" );
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document.write( "there is another way to analyze it.\r<BR>\n" );
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document.write( "the answer should be the same.\r<BR>\n" );
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document.write( "the probability of getting 3 aces and 2 kings would be:\r<BR>\n" );
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document.write( "4/52 * 3/51 * 2/50 * 4/49 * 3/48 * 5! / (3! * 2!).\r<BR>\n" );
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document.write( "use your calculator to get 9.234463016 * 10^-6.\r<BR>\n" );
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document.write( "the answers are the same.\r<BR>\n" );
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document.write( "the two methods are good.\r<BR>\n" );
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document.write( "c(n,x) is equal to n! / (x! * (n-x)!).\r<BR>\n" );
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document.write( "this is the combination formula for the number of ways of getting x things out of n where order is not important.\r<BR>\n" );
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document.write( "if you want to get 3 aces out of 4 possible, the formula becomes c(4,3) = 4! / (3! * 1!) = (4*3*2*1) / (3*2*1*1) = 4.\r<BR>\n" );
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document.write( "if you want to get 2 kings out of 4 possible, the formula becomes c(4,2) = 4! / (2! * 2!) = (4*3*2*1) / (2*1*2*1) = 6.\r<BR>\n" );
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document.write( "the number of possible combinations of 3 aces and 2 kings becomes 4 * 6 = 24.\r<BR>\n" );
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document.write( "the number of ways you can get 5 cards, regardless of what they are, out of 52 cards is c(52,5) = 52! / (5! * 47!) = (52*51*50*49*48*47!) / (5*4*3*2*1*47!)/\r<BR>\n" );
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document.write( "the 47! in the numerator and denominator cancel out and you are left with (52*51*50*49*48) / (5*4*3*2*1) which is equal to 2598960.\r<BR>\n" );
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document.write( "24/2598960 = 9.234463016 * 10^-6 as we calculated before.\r<BR>\n" );
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document.write( "the second method finds the probability of of getting them in a specific order and then multiplies by the number of ways you can arrange them.\r<BR>\n" );
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document.write( "(4/52 * 3/51 * 2/50) is the probability of getting the 3 aces.\r<BR>\n" );
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document.write( "(4/49 * 4/48) is the probability of getting the 2 kings once you've drawn the 3 aces.\r<BR>\n" );
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document.write( "(4/52 * 3/51 * 2/50 * 4/49 * 4/48) is the probability of getting the 3 aces and the 4 kings in that specific order (first the 3 aces and then the 2 kings).\r<BR>\n" );
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document.write( "these 5 cards can be arranged in 5! ways.\r<BR>\n" );
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document.write( "however, 3 of them are the same and 2 of them are the same, therefore, the number of ways they can be arranged becomes 5! / (3! * 2!).\r<BR>\n" );
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document.write( "you are not distinguishing between the aces or the kings in anyway, such as suit, therefore they are assumed to be the same cards.\r<BR>\n" );
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document.write( "an ace is an ace regardless if it's a spade or a heart.\r<BR>\n" );
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document.write( "both ways are valid and useful to know.\r<BR>\n" );
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document.write( "i usually try to solve these both ways as a check against each method.\r<BR>\n" );
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document.write( "if the results are different, i try to find out why.\r<BR>\n" );
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document.write( "i hope you found this helpful.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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