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.
The sum of the three numbers on the spinner is one.
The largest number you can get in two spins of this spinner is one.
If you spin the spinner twice and add, you get a sum of one about a quarter of the time.
If you spun the spinner a hundred times and added up all the numbers, you'd probably get somewhere near 40.
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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