document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=48479>Question 48479</A>: <I><SPAN STYLE=\"word-break: break-all;\"> When a single card is drawn from an ordinary 52 card deck, find the probability it will be the following;\r<BR>\n" );
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document.write( "A red Card<BR>\n" );
document.write( "An Ace<BR>\n" );
document.write( "A heart or a king<BR>\n" );
document.write( "A club, given it is black<BR>\n" );
document.write( "An ace or a card above a jack<BR>\n" );
document.write( "A queen, given it is a spade<BR>\n" );
document.write( "A 5, given it is a face card\r<BR>\n" );
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document.write( "Now assume that the two cards are drawn without replacement.  Find the probability of each of the following events.\r<BR>\n" );
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document.write( "Both are aces<BR>\n" );
document.write( "One is a king and one is an ace<BR>\n" );
document.write( "Neither one is an ace<BR>\n" );
document.write( "Both are red<BR>\n" );
document.write( "Both are face cards<BR>\n" );
document.write( "A face card<BR>\n" );
document.write( "A black 2 or 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 #32084 by </B><A HREF=/tutors/aboutme.mpl?userid=AnlytcPhil><B>AnlytcPhil(1806)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=AnlytcPhil><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=AnlytcPhil><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=32084\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <PRE STYLE=\"white-space: pre-wrap; white-space: -moz-pre-wrap; white-space: -pre-wrap; white-space: -o-pre-wrap; word-wrap: break-word;\"><font size = 4><b>When a single card is drawn from an ordinary 52 card deck, find \r\n" );
document.write( "the probability it will be the following; \r\n" );
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document.write( "A red Card:  There are 26 red cards out of 52 cards, so the \r\n" );
document.write( "probability is P(R) = 26/52 = 1/2\r\n" );
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document.write( "An Ace:  There are 4 aces out of 52 cards, so the probability is\r\n" );
document.write( "P(A) = 4/52 = 1/13\r\n" );
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document.write( "A heart or a king:  P(H or K) = P(H)+P(K)-P(H and K)\r\n" );
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document.write( "There are 13 hearts out of 52 cards, so P(H) = 13/52 = 1/4\r\n" );
document.write( "There are 4 kings out of 52 cards, so P(K) = 4/52 = 1/13\r\n" );
document.write( "There is 1 card which is a heart and a king (the king of \r\n" );
document.write( "hearts), so P(H and K) = 1/52, so\r\n" );
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document.write( "P(H or K) = P(H)+P(K)-P(H and K) = 1/4+1/13-1/52 = \r\n" );
document.write( "13/52+4/52-1/52 = 16/52 = 4/13\r\n" );
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document.write( "A club, given it is black: P(C|B) = P(C and B)/P(B)\r\n" );
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document.write( "There are 13 cards out of 52 which Clubs and Blacks so \r\n" );
document.write( "P(C and B) = 13/52 = 1/4\r\n" );
document.write( "There are 26 cards out of 52 which are black, so\r\n" );
document.write( "P(B) = 26/52 = 1/2\r\n" );
document.write( "P(C|B) = P(C and B)/P(B) = (1/4)/(1/2) = (1/4)(2/1) = 1/2\r\n" );
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document.write( "An ace or a card above a jack\r\n" );
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document.write( "Hmmm!  This will depend on whether and ace is considered \r\n" );
document.write( "higher or lower than a jack.  It is usually higher, but \r\n" );
document.write( "sometimes it's lower. I'll do both cases:\r\n" );
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document.write( "Case 1.  An ace is considered higher than a jack:\r\n" );
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document.write( "P(A or Above Jack) = P(A) + P(Above Jack) - P(A and Above Jack)\r\n" );
document.write( "There are 4 aces, so P(A) = 4/52 = 1/13 \r\n" );
document.write( "There are 4 queens, 4 kings and 4 aces which are above jack, so\r\n" );
document.write( "P(Above jack) = 12/52 = 3/13\r\n" );
document.write( "All 4 aces are above Jack so P(A and Above Jack) = 4/52 = 1/13, so\r\n" );
document.write( "P(A or Above Jack) = P(A) + P(Above Jack) - P(A and Above Jack) =\r\n" );
document.write( "1/13 + 3/13 - 1/13 = 3/13\r\n" );
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document.write( "Case 2.  An ace is considered lower than a jack:\r\n" );
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document.write( "P(A or Above Jack) = P(A) + P(Above Jack) - P(A and Above Jack)\r\n" );
document.write( "There are 4 aces, so P(A) = 4/52 = 1/13 \r\n" );
document.write( "There are 4 queens and 4 kings which are above jack, so\r\n" );
document.write( "P(Above jack) = 8/52 = 2/13\r\n" );
document.write( "None of the 4 aces are above Jack so P(A and Above Jack) = 0/52 =\r\n" );
document.write( "0, so \r\n" );
document.write( "P(A or Above Jack) = P(A) + P(Above Jack) - P(A and Above Jack) =\r\n" );
document.write( "1/13 + 3/13 - 0 = 4/13\r\n" );
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document.write( "A queen, given it is a spade: P(Q|S) = P(Q and S)/P(S)\r\n" );
document.write( "There is only one card that is a queen and a spade, namely the \r\n" );
document.write( "queen of spades, so P(Q and S) = 1/52. There are 13 spades, \r\n" );
document.write( "so P(S) = 13/52 = 1/4, so\r\n" );
document.write( "P(Q|S) = P(Q and S)/P(S) = (1/52)/(1/4) = (1/52)(4/1) = 1/13\r\n" );
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document.write( "A 5, given it is a face card\r\n" );
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document.write( "P(5|F) = P(5 and F)/P(F)\r\n" );
document.write( "Since a 5 is not a face card, P(5 and F) = 0\r\n" );
document.write( "Since there are 4 Jacks, 4 queens, and 4 kings which are face \r\n" );
document.write( "cards, P(F) = 12/52 = 3/13, so\r\n" );
document.write( "P(5|F) = P(5 and F)/P(F) = 0/(3/13) = 0(13/3) = 0\r\n" );
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document.write( "Now assume that the two cards are drawn without replacement. \r\n" );
document.write( "Find the probability of each of the following events. \r\n" );
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document.write( "Both are aces:\r\n" );
document.write( "P(A 1st and A 2nd) = P(A 1st)·P(A 2nd) = (4/52)(3/51) = 1/221 \r\n" );
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document.write( "One is a king and one is an ace:\r\n" );
document.write( "P[(A 1st and K 2nd) or (K 1st and A 2nd)] = \r\n" );
document.write( "P(A 1st and K 2nd) + P(K 1st and A 2nd) =\r\n" );
document.write( "P(A 1st)·P(K 2nd) + P(K 1st)·P(A 2nd) =\r\n" );
document.write( "(4/52)(4/51) + (4/52)(4/51) = 8/663\r\n" );
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document.write( "Neither one is an ace:\r\n" );
document.write( "There are 48 non-aces, so P(non-A 1st and non-A 2nd) =\r\n" );
document.write( "P(non-A 1st)·P(non-A 2nd) = (48/52)(47/51) = 188/221 \r\n" );
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document.write( "Both are red:\r\n" );
document.write( "P(R 1st and R 2nd) = P(R 1st)·P(R 2nd) = (26/52)(25/51) = \r\n" );
document.write( "25/102\r\n" );
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document.write( "Both are face cards:\r\n" );
document.write( "P(F 1st and F 2nd) = P(F 1st)·P(F 2nd) = (12/52)(11/51) = \r\n" );
document.write( "11/221\r\n" );
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document.write( "A face card\r\n" );
document.write( "Here it is easier to use the complement event, \r\n" );
document.write( "P(one is a face card) = 1 - P(both are non-face card). \r\n" );
document.write( "(There are 12 face cards and 40 non-face cards)\r\n" );
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document.write( "1 - P(non-F 1st and non-F 2nd) = 1 - P(non-F 1st)·P(non-F 2nd) =\r\n" );
document.write( "1 - (40/52)(39/51) = 7/17 \r\n" );
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document.write( "This could have also been worked a longer way without using the \r\n" );
document.write( "complement event:\r\n" );
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document.write( "P[(F 1st and F 2nd) or (F 1st and non-F 2nd) or (non-F 1st and F 2nd)] =\r\n" );
document.write( "P(F 1st and F 2nd) + P(F 1st and non-F 2nd) + P(non-F 1st and F 2nd) =\r\n" );
document.write( "P(F 1st)·P(F 2nd) + P(F 1st)·P(non-F 2nd) + P(non-F 1st)·P(F 2nd) =\r\n" );
document.write( "(12/52)(11/51) + (12/52)(40/51) + (40/52)(12/52) =\r\n" );
document.write( "11/221 + 40/221 + 40/221 = 7/17\r\n" );
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document.write( "A black 2 or a face card\r\n" );
document.write( "This one is also much easier to work using the complement event:\r\n" );
document.write( "There are 2 black 2's and 12 face cards or 14 which are one or the other.\r\n" );
document.write( "That leaves 38 cards which are neither.\r\n" );
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document.write( "P(B2 or F) = 1 - P(neither 1st and neither 2nd) =\r\n" );
document.write( "P(neither 1st)·P(neither 2nd) = (38/52)·(37/51) = 703/1326 \r\n" );
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document.write( "This could have also been done a much longer way without using the\r\n" );
document.write( "complement event, but I will not bother with that.\r\n" );
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document.write( "Edwin</PRE> <BR>\n" );
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