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Difference between revisions of "2007 AMC 8 Problems/Problem 25"
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One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is <math>3\pi + (2 \cdot 9\pi) = 21\pi</math>. Since the area of the whole circle is <math>36\pi</math>, the probability of getting an odd is <math>\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}</math>.
 
One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is <math>3\pi + (2 \cdot 9\pi) = 21\pi</math>. Since the area of the whole circle is <math>36\pi</math>, the probability of getting an odd is <math>\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}</math>.
  
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Since the remaining sections contain even numbers (that is, 2), the probability of throwing an odd is the complement, or <math>1 - \frac{7}{12} = \frac{5}{12}</math>.
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Since the remaining sections contain even numbers (that is, 2), the probability of throwing an even is the complement, or <math>1 - \frac{7}{12} = \frac{5}{12}</math>.
  
  

Revision as of 20:44, 5 October 2015

Problem

On the dart board shown in the Figure, the outer circle has radius $6$ and the inner circle has radius $3$. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd?

AMC8 2007 25.png $\mathrm{(A)} \frac{17}{36} \qquad \mathrm{(B)} \frac{35}{72} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{37}{72} \qquad \mathrm{(E)} \frac{19}{36}$

Solution

To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. Before we do that, we also have to figure out some relative areas. You could either find the absolute areas (as is done below for completeness), or apply some proportional reasoning and symmetry to realize the relative areas (which is really all that matters).


To find the areas of the sections, notice that the three smaller sections trisect a circle with radius $3$. The area of this entire circle is $9\pi$. The area of each smaller section then must be $\frac{9\pi}{3}$ or $3\pi$. The larger sections trisect an "ring" which is the difference of two circles, one with radius $3$, the other radius $6$. So, the area of the ring (annulus) is $36\pi - 9\pi$ or $27\pi$. The area of each larger section must be $\frac{27\pi}{3}$ or $9\pi$. Note that the area of the whole circle is $36\pi$.


One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is $3\pi + (2 \cdot 9\pi) = 21\pi$. Since the area of the whole circle is $36\pi$, the probability of getting an odd is $\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}$.

Since the remaining sections contain even numbers (that is, 2), the probability of throwing an even is the complement, or $1 - \frac{7}{12} = \frac{5}{12}$.


Now, the two cases: You could either get an odd then an even, or an even then an odd.


CASE 1: Odd then even Multiply the probabilities to get $\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}$.


CASE 2: Even then odd Multiply the probabilities to get $\frac{5}{12} \cdot \frac{7}{12} = \frac{35}{144}$. Notice that this is the same.


Thus, the total probability of an odd sum is $\frac{35}{144} \cdot \frac{2}{1}$ = $\boxed{B = \frac{35}{72}}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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