Difference between revisions of "2006 AMC 8 Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even + Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner <math>Q</math> contains only even numbers. Therefore we must work with Even + Even + Odd and spinner <math>Q</math> will give us one of our even numbers. We also see that spinner <math>R</math> only contains odd, so spinner <math>R</math> must give us our odd | + | In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even + Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner <math>Q</math> contains only even numbers. Therefore we must work with Even + Even + Odd and spinner <math>Q</math> will give us one of our even numbers. We also see that spinner <math>R</math> only contains odd, so spinner <math>R</math> must give us our odd number. We still need one even number from spinner <math>P</math>. There is only 1 even number: <math>2</math>. Since spinning the required numbers are automatic on the other spinners, we only have to find the probability of spinning a <math>2</math> in spinner <math>P</math>, which clearly is <math>\boxed{\textbf{(B)}\ \frac{1}{3}}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=16|num-a=18}} | {{AMC8 box|year=2006|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:08, 3 August 2017
Problem
Jeff rotates spinners ,
and
and adds the resulting numbers. What is the probability that his sum is an odd number?
Solution
In order for Jeff to have an odd number sum, the numbers must either be Odd + Odd + Odd or Even + Even + Odd. We easily notice that we cannot obtain Odd + Odd + Odd because spinner contains only even numbers. Therefore we must work with Even + Even + Odd and spinner
will give us one of our even numbers. We also see that spinner
only contains odd, so spinner
must give us our odd number. We still need one even number from spinner
. There is only 1 even number:
. Since spinning the required numbers are automatic on the other spinners, we only have to find the probability of spinning a
in spinner
, which clearly is
See Also
2006 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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