Difference between revisions of "2021 WSMO Speed Round Problems/Problem 9"
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The probability of a dart getting 70 points is <math>\frac{\pi}{196}</math>, the probability of a dart getting <math>-28</math> points is <math>\frac{8\pi}{196}</math>, the probability of a dart getting <math>49</math> points is <math>\frac{16\pi}{196}</math>, and the probability of a dart getting <math>-21</math> points is <math>\frac{24\pi}{196}</math>. This means that on a random throw, the expected number of points Bobby gets is <cmath>70 \cdot \frac{\pi}{196} - 28 \cdot \frac{8\pi}{196} + 49 \cdot \frac{16\pi}{196} - 21 \cdot \frac{24\pi}{196} = \frac{9}{14}\pi</cmath>. After 20 throws, Bobby is expected to get <cmath>20\cdot\frac{9}{14}\pi=\frac{90}{7}\pi\Longrightarrow90+7=\boxed{97}.</cmath> | The probability of a dart getting 70 points is <math>\frac{\pi}{196}</math>, the probability of a dart getting <math>-28</math> points is <math>\frac{8\pi}{196}</math>, the probability of a dart getting <math>49</math> points is <math>\frac{16\pi}{196}</math>, and the probability of a dart getting <math>-21</math> points is <math>\frac{24\pi}{196}</math>. This means that on a random throw, the expected number of points Bobby gets is <cmath>70 \cdot \frac{\pi}{196} - 28 \cdot \frac{8\pi}{196} + 49 \cdot \frac{16\pi}{196} - 21 \cdot \frac{24\pi}{196} = \frac{9}{14}\pi</cmath>. After 20 throws, Bobby is expected to get <cmath>20\cdot\frac{9}{14}\pi=\frac{90}{7}\pi\Longrightarrow90+7=\boxed{97}.</cmath> | ||
~pinkpig | ~pinkpig | ||
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Latest revision as of 09:08, 11 July 2022
Problem
Bobby is going to throw 20 darts at the dartboard shown below. It is formed by 4 concentric circles, with radii of 
and 
, with the largest circle being inscribed in a square. Each point on the dartboard has an equally likely chance of being hit by a dart, and Bobby is guaranteed to hit the dartboard. Each region is labeled with its point value (the number of points Bobby will get if he hits that region). The expected number of points Bobby will get after throwing the 20 darts can be expressed as 
where 
. Find 
.
![[asy] size(5cm); filldraw(shift(-7,-7)*((0,0)--(14,0)--(14,14)--(0,14)--cycle),gray); filldraw(circle((0,0),7),white); filldraw(circle((0,0),5),white); filldraw(circle((0,0),3),white); filldraw(circle((0,0),1),white); label("$70$",(0,0)); label("$-28$",(0,2)); label("$49$",(0,4)); label("$-21$",(0,6)); label("$0$",(-6,6)); label("$0$",(6,6)); label("$0$",(-6,-6)); label("$0$",(6,-6)); [/asy]](http://latex.artofproblemsolving.com/a/f/a/afa99ca5c6ecf34b940029025c5aee1edd26e0b1.png)
Solution
The probability of a dart getting 70 points is 
, the probability of a dart getting 
points is 
, the probability of a dart getting 
points is 
, and the probability of a dart getting 
points is 
. This means that on a random throw, the expected number of points Bobby gets is ![\[70 \cdot \frac{\pi}{196} - 28 \cdot \frac{8\pi}{196} + 49 \cdot \frac{16\pi}{196} - 21 \cdot \frac{24\pi}{196} = \frac{9}{14}\pi\]](http://latex.artofproblemsolving.com/4/9/9/4992111fa6aa3861e21c0f69875225f64e94acaa.png)
. After 20 throws, Bobby is expected to get ![\[20\cdot\frac{9}{14}\pi=\frac{90}{7}\pi\Longrightarrow90+7=\boxed{97}.\]](http://latex.artofproblemsolving.com/b/6/2/b62575ced5afb04f1bd9991f930dc5861de69f5b.png)
~pinkpig
