document.write( "Question 1183432: The faces of a fair cube are numbered 1 through 6; the probability of rolling any number form 1 through 6 is equally likely. If the cube is rolled twice, what is the probability that an even number will appear on the top face in the first roll or that the number 1 will appear on top face in the second roll? \n" ); document.write( "

Algebra.Com's Answer #813760 by math_helper(2461) $\"\"$ $\"About$

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document.write( "An easy way to solve this is to compute \r
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document.write( "    1 - P(complement of indicated outcome)\r
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document.write( "P(complement of indicated outcome) = P(odd on 1st roll) * P(2 thru 6 on 2nd roll)
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document.write( "= 3/6 * 5/6 = 15/36\r
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document.write( "and  1 - 15/36 = 21/36  or  \r
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document.write( "A longer way is to recognize that there is overlap between the two cases, and we can break it down to three non-overlapping outcomes:\r
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document.write( "P = P(odd on 1st roll) * P(1 on 2nd roll)
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document.write( "    + P(even on 1st roll) * P(2-6 on 2nd roll)
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document.write( "    + P(even on 1st roll) * P(1 on 2nd roll)\r
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document.write( "=  (3/6)*(1/6) + (3/6)*(5/6) + (3/6)*(1/6) = (3+15+3)/36 = 21/36 = 7/12, as above.
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document.write( "[ The third term is the case that BOTH rolls produce one of the desired
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document.write( "outcomes. ]\r
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document.write( "---\r
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document.write( "Yet another way is to use the inclusion-exclusion principle (we add the two probabilities but, recognizing we've overcounted, subtract out the \"common\" part which was counted twice):\r
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document.write( "   P = (even on 1st) + (1 on 2nd) - P(even on 1st AND 1 on 2nd)\r
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document.write( "   P = (3/6)+(1/6) - (3/6)*(1/6) = 4/6 - 3/36 =  24/36 - 3/36 = 21/36 = 7/12, as before.  \r
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