Posted by Daniel on Friday, January 9, 2009 at 12:25am.
I've worked out the first part of the question, just need help with the second part which is:
The spinner has three sections. While no two are the same size, one of the sections is half the size of another.
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You are more likely to get 2/3 than one if you sum two spins, but 5/6 is the most likely sum of all.
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The sum of the three numbers on the spinner is one.
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The largest number you can get in two spins of this spinner is one.
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If you spin the spinner twice and add, you get a sum of one about a quarter of the time.
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If you spun the spinner a hundred times and added up all the numbers, you'd probably get somewhere near 40.
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DRAW THE SPINNER.

Math (Probability)  economyst, Friday, January 9, 2009 at 9:37am
I don't think the spinner is possible given the requirements you provided.
The largest number in two spins is 1 so, one section must have a value of 1/2 (or 6/12)
The most likely value for two spins is 5/6 so, the largest area must have a value of 5/12.
Which means the value for the last area must be 1/12  which means there is no way to get a value of 2/3 on two spins.
I would like to see the correct answer.
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