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Difference between revisions of "2017 AMC 10A Problems/Problem 18"
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==Solution==
 
==Solution==
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Let <math>P</math> be the probability Amelia wins. Note that <math>P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P</math>, as if she gets to her turn again, she is back where she started with probability of winning <math>P</math>. The chance she wins on her first turn is <math>\frac{1}{3}</math>, and the chance she makes it to her turn again is a combination of her failing to win the first turn—<math>\frac{2}{3}</math> and Blaine failing to win—<math>\frac{3}{5}</math>. Multiplying gives us <math>\frac{2}{5}</math>. Thus,
+
Let <math>P</math> be the probability Amelia wins. Note that <math>P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P</math>, as if she gets to her turn again, she is back where she started with probability of winning <math>P</math>. The chance she wins on her first turn is <math>\frac{1}{3}</math>. The chance she makes it to her turn again is a combination of her failing to win the first turn—<math>\frac{2}{3}</math> and Blaine failing to win—<math>\frac{3}{5}</math>. Multiplying gives us <math>\frac{2}{5}</math>. Thus,
−
<cmath>P = \frac{1}{3} + \frac{2}{5} \implies P = frac{5}{9}</cmath>.
+
<cmath>P = \frac{1}{3} + \frac{2}{5} \implies P = \frac{5}{9}</cmath>
 
Finally, we do <math>9-5=\boxed{\textbf{(E)}\ 4}</math>.
 
Finally, we do <math>9-5=\boxed{\textbf{(E)}\ 4}</math>.

Revision as of 17:20, 8 February 2017

Problem

Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $q-p$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P$, as if she gets to her turn again, she is back where she started with probability of winning $P$. The chance she wins on her first turn is $\frac{1}{3}$. The chance she makes it to her turn again is a combination of her failing to win the first turn—$\frac{2}{3}$ and Blaine failing to win—$\frac{3}{5}$. Multiplying gives us $\frac{2}{5}$. Thus, \[P = \frac{1}{3} + \frac{2}{5} \implies P = \frac{5}{9}\] Finally, we do $9-5=\boxed{\textbf{(E)}\ 4}$.

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