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.

•

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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