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

## 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 $\frac{1}{3}$, and 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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