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Difference between revisions of "2007 AMC 8 Problems/Problem 25"

(Created page with 'In our diagram, the three smaller sections trisect a circle with radius <math>3</math>. Thus, the area of the entire section is <math>9\pi</math>. The area of each section then m…')
 
 
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In our diagram, the three smaller sections trisect a circle with radius <math>3</math>. Thus, the area of the entire section is <math>9\pi</math>. The area of each section then must be <math>\frac{9\pi}{3}</math> or <math>3\pi</math>. The area of the larger regions trisect an area which is the difference of two circles, one with radius <math>3</math>, the other radius <math>6</math>. So, the area of the region is <math>36\pi - 9\pi</math> or <math>27\pi</math>. The area of each section must be <math>\frac{27\pi}{3}</math> or <math>9\pi</math>. 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.
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==Problem==
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On the dart board shown in the figure below, the outer circle has radius <math>6</math> and the inner circle has radius <math>3</math>. 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 of the regions hit. What is the probability that the score is odd?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
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<center><asy>
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draw(circle((0,0),6));
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draw(circle((0,0),3));
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draw((0,0)--(0,6));
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draw((0,0)--(rotate(120)*(0,6)));
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draw((0,0)--(rotate(-120)*(0,6)));
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label('1',(rotate(60)*(0,3/2)));
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label('2',(rotate(-60)*(0,3/2)));
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label('2',(0,-3/2));
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label('2',(rotate(60)*(0,9/2)));
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label('1',(rotate(-60)*(0,9/2)));
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label('1',(0,-9/2));
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</asy></center>
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<math>\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}</math>
  
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CASE 1:
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==Solution 1==
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On the first throw, you will get an odd number (<math>1</math>). Only one <math>3\pi</math> section holds a <math>1</math>, but two <math>9\pi</math> sections hold a <math>1</math>, so, <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>. Since all that is left is even numbers, the probability of getting an even is <math>1 - \text{ pobability of getting an odd }</math> (do you see why). So it is <math>\frac{12}{12} - \frac{7}{12} = \frac{5}{12}</math>. So, we now multiply both probabilities to get <math>\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}</math>.
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To get an odd sum, we must add an even number and an odd number (a 1 and a 2). 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).
  
  
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CASE 2:
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The area of this entire smaller circle is <math>9\pi</math>. The area of each smaller section then must be <math>\frac{9\pi}{3}</math> or <math>3\pi</math>. The larger sections trisect a "ring" which is the difference of two circles, one with radius <math>3</math>, the other radius <math>6</math>. So, the area of the ring (''annulus'') is <math>36\pi - 9\pi</math> or <math>27\pi</math>. The area of each larger section must be <math>\frac{27\pi}{3}</math> or <math>9\pi</math>. Note that the area of the whole circle is <math>36\pi</math>.
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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. (<math>e</math>) and multiply that by odd prob. (<math>o</math>), which is exactly what we just did! So this probability is just <math>\frac{35}{144}</math>.  
 
  
  
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Instead of adding (which we could do), since they are the same number we can just multiply by <math>2</math> (n + n does equal 2 *n). We will then get <math>\frac{35}{144} \cdot \frac{2}{1} = \text{ ( cross cancel ) } = \frac{35}{144 \div 2} = \boxed{\frac{35}{72} \text{ which is } \bold{(B)}}</math>.
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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>.
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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>.
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Now, the two cases: You could either get an odd then an even, or an even then an odd.
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 +
 
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Case 1: Odd, then even.
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Multiply the probabilities to get <math>\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}</math>.
 +
 
 +
 
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Case 2: Even, then odd.
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Multiply the probabilities to get <math>\frac{5}{12} \cdot \frac{7}{12} = \frac{35}{144}</math>. Notice that this is the same.
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 +
 
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Thus, the total probability of an odd sum is <math>\frac{35}{144} \cdot \frac{2}{1} = \boxed{\textbf{(B)} = \frac{35}{72}}</math>.
 +
 
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==Video Solution 1==
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https://www.youtube.com/watch?v=93RsgRTJYUM
 +
 
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== Video Solution by OmegaLearn ==
 +
https://youtu.be/tKsYSBdeVuw?t=8
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/iFSbbM-suKA
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2007|num-b=24| after=Last problem}}
 +
{{MAA Notice}}

Latest revision as of 13:18, 29 October 2024

Problem

On the dart board shown in the figure below, 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 of the regions hit. What is the probability that the score is odd?

[asy] draw(circle((0,0),6)); draw(circle((0,0),3)); draw((0,0)--(0,6)); draw((0,0)--(rotate(120)*(0,6))); draw((0,0)--(rotate(-120)*(0,6))); label('1',(rotate(60)*(0,3/2))); label('2',(rotate(-60)*(0,3/2))); label('2',(0,-3/2)); label('2',(rotate(60)*(0,9/2))); label('1',(rotate(-60)*(0,9/2))); label('1',(0,-9/2)); [/asy]

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

Solution 1

To get an odd sum, we must add an even number and an odd number (a 1 and a 2). 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).


The area of this entire smaller circle is $9\pi$. The area of each smaller section then must be $\frac{9\pi}{3}$ or $3\pi$. The larger sections trisect a "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 $\frac{35}{144} \cdot \frac{2}{1} = \boxed{\textbf{(B)} = \frac{35}{72}}$.

Video Solution 1

https://www.youtube.com/watch?v=93RsgRTJYUM

Video Solution by OmegaLearn

https://youtu.be/tKsYSBdeVuw?t=8

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/iFSbbM-suKA

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last problem
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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