document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1035442>Question 1035442</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An ordinary (fair) dice is a cube with the numbers 1 through 6 on the sides (represented by painted spots). A dice is rolled twice in succession and that the face values of the the rolls are added together. The sum is recorded as the outcome of a single trial of a random experiment. Compute the probability of each of the following events:  event A: the sum is greater than 6 <BR>\n" );
document.write( "Event B : the sum is not divisible by 3 and not divisible by 6. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #650128 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=650128\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> before you can determine the probabilities, you have to analyze all the possible outcomes.\r<BR>\n" );
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document.write( "you have 36 possible outcomes, because 6 * 6 = 36.\r<BR>\n" );
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document.write( "the probability of getting a 2 is 1+1 = 1/36\r<BR>\n" );
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document.write( "the probability of getting a 3 is 1+2 or 2+1 = 2/36\r<BR>\n" );
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document.write( "the probability of getting a 4 is 1+3 or 2+2 or 3+1 = 3/36\r<BR>\n" );
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document.write( "the probability of getting a 5 is 1+4 or 2+3 or 3+2 or 4+1 = 4/36\r<BR>\n" );
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document.write( "the probability of getting a 6 is 1+5 or 2+4 or 3+3 or 4+2 or 5+1 = 5/36\r<BR>\n" );
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document.write( "the probability of getting a 7 is 1+6 or 2+5 or 3+4 or 4+3 or 5+2 or 6+1 = 6/36\r<BR>\n" );
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document.write( "the probability of getting an 8 is 2+6 or 3+5 or 4+4 or 5+3 or 6+2 = 5/36\r<BR>\n" );
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document.write( "the probability of getting a 9 is 3+6 or 4+5 or 5+4 or 6+3 = 4/36\r<BR>\n" );
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document.write( "the probability of getting a 10 is 4+6 or 5+5 or 6+4 =3/36\r<BR>\n" );
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document.write( "the probability of getting an 11 is 5+6 or 6+5 = 2/36\r<BR>\n" );
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document.write( "the probability of getting a 12 is 6+6 = 1/36.\r<BR>\n" );
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document.write( "add up the total probabilities and you get (1+2+3+4+5+6+5+4+3+2+1)/36 = 36/36.\r<BR>\n" );
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document.write( "that's equivalent to 1, as it should be since the sum of all probabilities has to be equal to 1.\r<BR>\n" );
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document.write( "now that you know what the probabilities are, you can figure out the solution to the problem.\r<BR>\n" );
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document.write( "the problem says:\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 6.\r<BR>\n" );
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document.write( "the probability that the sum is greater than 6 is equal to the probability that the sum is 7 or 8 or 9 or 10 or 11 or 12.<BR>\n" );
document.write( "this is equal to (6 + 5 + 4 + 3 + 2 + 1) / 36 = 21/36.\r<BR>\n" );
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document.write( "if this is correct, then the probability that the sum is less than or equal to 6 must be equal to 36/36 - 21/36 = 15/36.\r<BR>\n" );
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document.write( "the probability that the sum is less than or equal to 6 is equal to (1 + 2 + 3 + 4 + 5) / 36 = 15/36, so it checks out ok.\r<BR>\n" );
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document.write( "Event B : the sum is not divisible by 3 and not divisible by 6.\r<BR>\n" );
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document.write( "if the number is divisible by 6, then it is also divisible by 3, so all you need to do is find the numbers that are divisible by 3.<BR>\n" );
document.write( "the possible numbers are 2,3,4,5,6,7,8,9,10,11,12<BR>\n" );
document.write( "of these numbers, take out the numbers that are divisible by 3 to get:<BR>\n" );
document.write( "2,4,5,7,8,10,11.<BR>\n" );
document.write( "the sum of the probabilities of rolling these numbers would be:<BR>\n" );
document.write( "(1 + 3 + 4 + 6 + 5 + 3 + 2) / 36 = 24/36.\r<BR>\n" );
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document.write( "if this is correct, then the probability of the number being divisible by 3 or 6 would have to be 36/36 - 24/36 = 12/36.\r<BR>\n" );
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document.write( "that would be the probability of getting a 3 or 6 or 9 or 12.\r<BR>\n" );
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document.write( "the probability of that is (2 + 5 + 4 + 1) / 36 = 12/36, so it checks out ok.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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