2017 AMC 10A Problems/Problem 18

Revision as of 19:36, 8 February 2017 by Thedoge (talk | contribs) (Solution)


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 $q-p$?

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


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, \[P = \frac{1}{3} + \frac{2}{5}P\] Therefore, $P = \frac{5}{9}$, so the answer is $9-5=\boxed{\textbf{(D)}\ 4}$.

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions

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