Question 238130
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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.  


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


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.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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