document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1118623>Question 1118623</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 2.	An experiment consist of rolling a pair of dice. Event E1 is the event that 7 occurs, I.e., the sum on the dice is 7. Event E2 is the event that an odd number occurs on die 1. Find <BR>\n" );
document.write( "a.	Pr {E1}<BR>\n" );
document.write( "b.	Pr{E2}<BR>\n" );
document.write( "c.	Pr{E1&#8745;E2}<BR>\n" );
document.write( "d.	Pr{E1UE2}<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #734000 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=734000\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a)  Pr{E1} = 6/36 = <b>1/6</b>  (there are 6 ways to roll a 7 out of 36 possible outcomes)<br><BR>\n" );
document.write( "b)  Pr{E2} = 3/6  = <b>1/2</b>  (there are 3 ways to roll an odd number out of 6 possible outcomes)<br><BR>\n" );
document.write( "c)  Pr{E1 &#8745; E2} = 3/36 = <b>1/12</b>  (3 out of 36 possible outcomes fall in the intersection, or, one can multiply the results of (a) and (b))<br><BR>\n" );
document.write( "d)  Pr{E1 U E2 } = 21/36 = <b>7/12</b>   (this is the set of elements that comprise E1, E2, or both E1 & E2.  We can count them: E1: Die1={1, 3, 5} has 18 outcomes, and E2 {sum of Die1 and Die2 = 7} has 6 outcomes, but we've over counted by 3 outcomes (where Die1 is odd and Die1+Die2 = 7).  One can also compute (a) + (b) - (c) = 6/36+18/36 - 3/36 = 21/36 = 7/12. \r<BR>\n" );
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