document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1062196>Question 1062196</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose you randomly draw a single card from a standard deck of cards.<BR>\n" );
document.write( "a. what is the probability you will draw a six of diamonds?<BR>\n" );
document.write( "b. what is the probability you will draw a club?<BR>\n" );
document.write( "2. Suppose you randomly draw two cards from a standard deck of playing cards. What is the probability of of drawing a four of clubs and then drawing a seven of hearts? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #677044 by </B><A HREF=/tutors/aboutme.mpl?userid=math_helper><B>math_helper(2461)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_helper><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=math_helper><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=677044\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a.  There are 52 cards in the deck, and the 6 of diamonds is just one of them.  So Pr{drawing 6 diamonds} = 1/52<BR>\n" );
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document.write( "b.  There are 13 clubs in a deck of cards, so Pr{drawing a club} is 13/52 = 1/4<BR>\n" );
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document.write( "c.  If you randomly chose a card, you have 1/52 chance of that being the exact card you want.  If you then draw a 2nd card there is a 1/51 chance of it being the 2nd card you hoped to see.  I assume no replacement.  Thus, Pr{draw 4 of clubs followed by draw 7 of hearts} = (1/52)(1/51) = 1/2652. </SPAN>\n" );
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