document.write( "Question 238130: a six sided die is weighed so that the probability of \"4\" is 4 times as likely as getting any other number. what is the probability of getting a \"1\" or a \"3\"? \n" ); document.write( "

Algebra.Com's Answer #175007 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "A fair 6-sided die would have a 1 in 6 probability of a 4. But this die has a 4 of something probability when each of the other possibilities has a 1 in something probability. Since the numerator of the \"get a 4\" probability is 4 and the other 5 are 1, we have a denominator of 9: 4 plus 5 times 1 = 9. \r
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\n" ); document.write( "\n" ); document.write( "The probability of getting a 1 or a 3 is the sum of the probability of getting a 1 plus the probability of getting a 3. 1/9 + 1/9 = 2/9\r
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\n" ); document.write( "\n" ); document.write( "The only thing that bothers me about this problem is how someone would weight a die so that the 4 was equally more likely than any of the other numbers. To make the 4 come up more often, you would have to weight the side with the 3, certainly making the 4 more likely, but also making the 3 somewhat less likely than the 1, 2, 5, or 6. I guess this is the same as \"neglecting air resistance\" when you are asked to calculate the height of a thrown rock.\r
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\n" ); document.write( "\n" ); document.write( "John
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