document.write( "Question 1194146: 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
\n" ); document.write( "\n" ); document.write( "Compute the probability of each of the following events:
\n" ); document.write( "Event A: the sum is greater than 8.
\n" ); document.write( "Event B: the sum is an even number.
\n" ); document.write( "round answer two demical points
\n" ); document.write( "P(A)=
\n" ); document.write( "P(B)= \n" ); document.write( "
Algebra.Com's Answer #826268 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "We'll start with P(A).\r
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\n" ); document.write( "\n" ); document.write( "Let's say we have two dice, one red and the other blue
\n" ); document.write( "Each is labeled 1 through 6
\n" ); document.write( "The 36 possible dice sums are listed below
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+123456
1234567
2345678
3456789
45678910
567891011
6789101112
Use that table to determine...
  • There are 4 ways to roll a 9
  • There are 3 ways to roll a 10
  • There are 2 ways to roll an 11
  • There is 1 way to roll a 12
In all, there are 4+3+2+1 = 10 ways to get a sum that is larger than 8.
\n" ); document.write( "This is out of the 36 total outcomes
\n" ); document.write( "10/36 = (5*2)/(18*2) = 5/18 = 0.28 is the approximate value of P(A)
\n" ); document.write( "Event A happens roughly 28% of the time. \r
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\n" ); document.write( "\n" ); document.write( "Now onto calculating P(B)\r
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\n" ); document.write( "\n" ); document.write( "Refer to the chart above
  • There is 1 way to roll a 2
  • There are 3 ways to roll a 4
  • There are 5 ways to roll a 6
  • There are 5 ways to roll an 8
  • There are 3 ways to roll a 10
  • There is 1 way to roll a 12
Adding up those frequencies gets us 1+3+5+5+3+1 = 18\r
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\n" ); document.write( "\n" ); document.write( "Then dividing that over the total 36 outcomes yields 18/36 = 1/2 = 0.50 which is the value of P(B).
\n" ); document.write( "This is to be expected since half of the numbers on the die are even, so the sum being even should happen half the time. \r
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\n" ); document.write( "\n" ); document.write( "Side notes:
\n" ); document.write( "even + even = even
\n" ); document.write( "odd + even = odd
\n" ); document.write( "odd + odd = even\r
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\n" ); document.write( "\n" ); document.write( "Answers:
\n" ); document.write( "P(A) = 0.28
\n" ); document.write( "P(B) = 0.50
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