document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=381230>Question 381230</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose one card is selected at random from an ordinary deck of 52 playing cards. Let\r<BR>\n" );
document.write( "\n" );
document.write( "A= event a queen is selected<BR>\n" );
document.write( "B= event a diamond is selected\r<BR>\n" );
document.write( "\n" );
document.write( "Determine P(B/A) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #270425 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=270425\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We're given that event A has occurred, which means that we know that a queen has been selected. There are 4 queens (one of each suit) and only one queen of diamonds. So P(B | A) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F4 ALIGN=MIDDLE  ALT=\"1%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Formally, P(B | A) = P(A intersect B)/P(A), so this means that P(B | A) = P(Queen and diamond)/P(Queen) = (1/52)/(4/52) = 1/4 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );