document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=137573>Question 137573</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help on this question. \r<BR>\n" );
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document.write( "An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.\r<BR>\n" );
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document.write( "Compute the probability of each of the following events: \r<BR>\n" );
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document.write( "Event A : The sum is greater than 7 .<BR>\n" );
document.write( "Event B : The sum is an odd number. <BR>\n" );
document.write( "Write your answers as exact fractions for <BR>\n" );
document.write( "P(A)=<BR>\n" );
document.write( "P(B)= </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #100639 by </B><A HREF=/tutors/aboutme.mpl?userid=Fombitz><B>Fombitz(32388)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Fombitz><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=Fombitz><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=100639\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Event A : The sum of two dice<BR>\n" );
document.write( "First look at all of the possible outcomes.<BR>\n" );
document.write( "First number rolled, second number rolled, then the sum.<BR>\n" );
document.write( "1 1=2<BR>\n" );
document.write( "1 2=3<BR>\n" );
document.write( "1 3=4<BR>\n" );
document.write( "1 4=5<BR>\n" );
document.write( "1 5=6<BR>\n" );
document.write( "1 6=7<BR>\n" );
document.write( "2 1=3<BR>\n" );
document.write( "2 2=4<BR>\n" );
document.write( "2 3=5<BR>\n" );
document.write( ".<BR>\n" );
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document.write( "6 1=7<BR>\n" );
document.write( "6 2=8<BR>\n" );
document.write( "6 3=9<BR>\n" );
document.write( "6 4=10<BR>\n" );
document.write( "6 5=11<BR>\n" );
document.write( "6 6=12<BR>\n" );
document.write( "All sums from 2 to 12 are possible out of 36 possible outcomes, with the following probabilities.<BR>\n" );
document.write( "2=1/36<BR>\n" );
document.write( "3=2/36<BR>\n" );
document.write( "4=3/36<BR>\n" );
document.write( "5=4/36<BR>\n" );
document.write( "6=5/36<BR>\n" );
document.write( "7=6/36<BR>\n" );
document.write( "8=5/36<BR>\n" );
document.write( "9=4/36<BR>\n" );
document.write( "10=3/36<BR>\n" );
document.write( "11=2/36<BR>\n" );
document.write( "12=1/36<BR>\n" );
document.write( "For the probability of the sum being greater than 7, you would add the probabilities of getting 8, 9, 10, 11, and 12.<BR>\n" );
document.write( "P(A)=(5+4+3+2+1)/36=15/36 or 5/18.<BR>\n" );
document.write( "Event B : The sum is an odd number. <BR>\n" );
document.write( "Add up all of the probabilities for the sum to be 3, 5, 7, 9, and 11.<BR>\n" );
document.write( "P(B)=(2+4+6+4+2)/36=18/36 or 1/2. <BR>\n" );
document.write( "Since half of the sums are even and half of the sums are odd, you could have come to the same conclusion. <BR>\n" );
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