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 $\frac{1}{3}$ . 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, $P = \frac{1}{3} + \frac{2}{5}P \implies P = \frac{5}{9}$ Finally, we do $9-5=\boxed{\textbf{(D)}\ 4}$ .

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2017 AMC 10A Problems/Problem 18

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