document.write( "Question 1072315: Assume that two fair dice are rolled. First compute P(F) and then P(F|E). Explain why one would expect the probability of F to change as it did when we added the condition that E had occurred.
\n" ); document.write( "F: the total is nine
\n" ); document.write( "E: an odd total shows on the dice
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Algebra.Com's Answer #687222 by addingup(3677) $\"\"$ $\"About$

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Here's all your possible outcomes:
\n" ); document.write( "(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
\n" ); document.write( "(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
\n" ); document.write( "(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
\n" ); document.write( "(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
\n" ); document.write( "(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
\n" ); document.write( "(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
\n" ); document.write( ":
\n" ); document.write( "Let's call one die a and the other b. There are 6 possible outcomes for a and 6 for b, and 6*6 = 36 Now we have a short cut, multiply the possible outcomes for a and b and you're done.
\n" ); document.write( "Note that I laid out six rows and six columns, 6*6 = 36 This is the total number of possible outcomes, the total is called the sample space. Our sample space is 36.
\n" ); document.write( ":
\n" ); document.write( "Each subset (1,1)(1,2)etc. is called an event
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\n" ); document.write( "Now note how many add up to 9: 3,6; 6,3; 4,5; 5,4. These are the possible events we are looking for, for us this will be P(E), the Probability of one of these Events happening.
\n" ); document.write( "P(E) = 4/36 = 1=9 \n" ); document.write( "

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