# Difference between revisions of "2017 AMC 10A Problems/Problem 18"

(Created page with "==Problem== Amelia has a coin that lands heads with probability <math>\frac{1}{3}</math>, and Blaine has a coin that lands on heads with probability <math>\frac{2}{5}</math>....") |
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Amelia has a coin that lands heads with probability <math>\frac{1}{3}</math>, and Blaine has a coin that lands on heads with probability <math>\frac{2}{5}</math>. 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 <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>q-p</math>? | Amelia has a coin that lands heads with probability <math>\frac{1}{3}</math>, and Blaine has a coin that lands on heads with probability <math>\frac{2}{5}</math>. 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 <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>q-p</math>? | ||

− | <math>\ | + | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> |

+ | |||

+ | ==Solution== | ||

+ | 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, | ||

+ | <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>. |

## Revision as of 18:20, 8 February 2017

## Problem

Amelia has a coin that lands heads with probability , and Blaine has a coin that lands on heads with probability . 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 , where and are relatively prime positive integers. What is ?

## Solution

Let be the probability Amelia wins. Note that , as if she gets to her turn again, she is back where she started with probability of winning . The chance she wins on her first turn is , and the chance she makes it to her turn again is a combination of her failing to win the first turn— and Blaine failing to win—. Multiplying gives us . Thus, . Finally, we do .