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Difference between revisions of "2024 SSMO Accuracy Round Problems/Problem 10"

(Created page with "==Problem== Bobby is spinning a rigged wheel with three sections labeled <math>A,B,</math> and <math>C.</math> Two integers <math>0\leq x,y\leq 50</math> are chosen randomly...")
 
 
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==Solution==
 
==Solution==
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For each <math>x,y,</math> the probability of the first spin being <math>C</math> is <math>\left(1-\frac{x+y}{100}\right)</math> and the probability of the first and second spins both land on <math>C</math> is <math>\left(1-\frac{x+y}{100}\right)^2.</math> So, the answer is <cmath>\frac{\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right)^2}{\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right)}.</cmath> First, <cmath>\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right) = 51^2\cdot \mathbb{E}\left(1-\frac{x+y}{100}\right) = 51^2\left(1-\mathbb{E}\left(\frac{x}{100}\right)-\mathbb{E}\left(\frac{y}{100}\right)\right)=</cmath><cmath>51^2\left(1-\frac{1}{4}-\frac{1}{4}\right)= \frac{51^2}{2}.</cmath> Now, <cmath>\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right)^2 = \sum_{0\le x,y,\le 50}1-2\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right)+\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right)^2.</cmath> We have <cmath>\sum_{0\le x,y,\le 50}1 = 51^2</cmath> and <cmath>2\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right) = 2\cdot51^2\cdot \mathbb{E}\left(\frac{x+y}{100}\right) = 2\cdot51^2\cdot \left(\mathbb{E}\left(\frac{x}{100}\right)+\left(\frac{y}{100}\right)\right) </cmath><cmath>= 51^2\cdot\left(\frac{1}{4}+\frac{1}{4}\right) = 2\cdot\left(\frac{51^2}{2}\right) = 51^2.</cmath> So, <cmath>\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right)^2 = 51^2-51^2+\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right)^2=\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right)^2.</cmath>
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We compute
 +
<cmath>\begin{align*}
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\sum_{0\le x,y,\le 50}\left(x+y\right)^2&=\sum_{0\le x,y,\le 50}(x^2+y^2)+2\sum_{0\le x,y,\le 50}xy\\
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&=2\cdot51^2\left(\mathbb{E}
 +
(x^2)\right)+2\left(\sum_{0\le x\le 50}x\right)^2\\
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&=2\cdot51^2\left(\frac{\sum_{i=0}^{50}i^2}{51}\right)+2\left(\frac{50\cdot51}{2}\right)^2\\
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&=\frac{51\cdot50\cdot51\cdot101}{3}+\frac{50\cdot51\cdot50\cdot51}{2}\\
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&=50\cdot51\cdot51\left(\frac{101}{3}+\frac{50}{2}\right)\\
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&=\frac{50\cdot51\cdot51\cdot176}{3}.
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\end{align*}</cmath>
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So, the answer is <cmath>\frac{\frac{\frac{50\cdot51\cdot51\cdot176}{3}}{100\cdot100}}{\frac{51\cdot51}{2}} = \frac{2\cdot50\cdot176}{3\cdot100\cdot100} = \frac{44}{75}\implies44+75 = \boxed{119}.</cmath>
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 +
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~SMO_Team

Latest revision as of 14:33, 10 September 2025

Problem

Bobby is spinning a rigged wheel with three sections labeled $A,B,$ and $C.$ Two integers $0\leq x,y\leq 50$ are chosen randomly and independently, such that there is a $x\%$ chance the wheel lands on $A$ and a $y\%$ chance the wheel lands on $B.$ Given that the wheel lands on $C$ the first time, the probability that it will land on $C$ if Bobby spins it again can be expressed as $\frac{m}{n},$ for relatively prime positive integers $m$ and $n.$ Find $m+n.$

Solution

For each $x,y,$ the probability of the first spin being $C$ is $\left(1-\frac{x+y}{100}\right)$ and the probability of the first and second spins both land on $C$ is $\left(1-\frac{x+y}{100}\right)^2.$ So, the answer is \[\frac{\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right)^2}{\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right)}.\] First, \[\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right) = 51^2\cdot \mathbb{E}\left(1-\frac{x+y}{100}\right) = 51^2\left(1-\mathbb{E}\left(\frac{x}{100}\right)-\mathbb{E}\left(\frac{y}{100}\right)\right)=\]\[51^2\left(1-\frac{1}{4}-\frac{1}{4}\right)= \frac{51^2}{2}.\] Now, \[\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right)^2 = \sum_{0\le x,y,\le 50}1-2\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right)+\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right)^2.\] We have \[\sum_{0\le x,y,\le 50}1 = 51^2\] and \[2\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right) = 2\cdot51^2\cdot \mathbb{E}\left(\frac{x+y}{100}\right) = 2\cdot51^2\cdot \left(\mathbb{E}\left(\frac{x}{100}\right)+\left(\frac{y}{100}\right)\right)\]\[= 51^2\cdot\left(\frac{1}{4}+\frac{1}{4}\right) = 2\cdot\left(\frac{51^2}{2}\right) = 51^2.\] So, \[\sum_{0\le x,y,\le 50}\left(1-\frac{x+y}{100}\right)^2 = 51^2-51^2+\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right)^2=\sum_{0\le x,y,\le 50}\left(\frac{x+y}{100}\right)^2.\] We compute \begin{align*} \sum_{0\le x,y,\le 50}\left(x+y\right)^2&=\sum_{0\le x,y,\le 50}(x^2+y^2)+2\sum_{0\le x,y,\le 50}xy\\ &=2\cdot51^2\left(\mathbb{E} (x^2)\right)+2\left(\sum_{0\le x\le 50}x\right)^2\\ &=2\cdot51^2\left(\frac{\sum_{i=0}^{50}i^2}{51}\right)+2\left(\frac{50\cdot51}{2}\right)^2\\ &=\frac{51\cdot50\cdot51\cdot101}{3}+\frac{50\cdot51\cdot50\cdot51}{2}\\ &=50\cdot51\cdot51\left(\frac{101}{3}+\frac{50}{2}\right)\\ &=\frac{50\cdot51\cdot51\cdot176}{3}. \end{align*} So, the answer is \[\frac{\frac{\frac{50\cdot51\cdot51\cdot176}{3}}{100\cdot100}}{\frac{51\cdot51}{2}} = \frac{2\cdot50\cdot176}{3\cdot100\cdot100} = \frac{44}{75}\implies44+75 = \boxed{119}.\]


~SMO_Team

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