2007 AMC 8 Problems/Problem 25

In our diagram, the three smaller sections trisect a circle with radius $3$ . Thus, the area of the entire section is $9\pi$ . The area of each section then must be $\frac{9\pi}{3}$ or $3\pi$ . The area of the larger regions trisect an area which is the difference of two circles, one with radius $3$ , the other radius $6$ . So, the area of the region is $36\pi - 9\pi$ or $27\pi$ . The area of each section must be $\frac{27\pi}{3}$ or $9\pi$ . To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. You could either get an odd number then an even number, or an even number then an odd number.

CASE 1: On the first throw, you will get an odd number ( $1$ ). Only one $3\pi$ section holds a $1$ , but two $9\pi$ sections hold a $1$ , so, $3\pi + (2 \cdot 9\pi) = 21\pi$ . 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 all that is left is even numbers, the probability of getting an even is $1 - \text{ pobability of getting an odd }$ (do you see why). So it is $\frac{12}{12} - \frac{7}{12} = \frac{5}{12}$ . So, we now multiply both probabilities to get $\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}$ .

CASE 2: This time you will get an even then an odd number. But this will be the same thing as an odd then an even, because first we will find the even prob. ( $e$ ) and multiply that by odd prob. ( $o$ ), which is exactly what we just did! So this probability is just $\frac{35}{144}$ .

Instead of adding (which we could do), since they are the same number we can just multiply by $2$ (n + n does equal 2 *n). We will then get $\frac{35}{144} \cdot \frac{2}{1} = \text{ ( cross cancel ) } = \frac{35}{144 \div 2} = \boxed{\frac{35}{72} \text{ which is } \bold{(B)}}$ .

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2007 AMC 8 Problems/Problem 25