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

Revision as of 20:00, 1 December 2020

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 1

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 $9-5=\boxed{\textbf{(D)}\ 4}$ .

Solution 2

Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} ...$ This can be represented by an infinite geometric series: $P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}.$ Therefore, $P = \frac{5}{9}$ , so the answer is $9-5 = \boxed{\textbf{(D)}\ 4}.$

Solution by ktong

~minor LaTeX edit by virjoy2001

Video Solution

https://www.youtube.com/watch?v=umr2Aj9ViOA

@@ Line 10: / Line 10: @@
 ==Solution 2==
-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 second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} ...</math>This can be represented by an infinite geometric series, <cmath>P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}</cmath>.
+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 second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} ...</math>This can be represented by an infinite geometric series: <cmath>P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}.</cmath>
-Therefore, <math>P = \frac{5}{9}</math>, so the answer is <math>9-5 = \boxed{\textbf{(D)}\ 4}</math>
+Therefore, <math>P = \frac{5}{9}</math>, so the answer is <math>9-5 = \boxed{\textbf{(D)}\ 4}.</math>
 Solution by ktong
+~minor LaTeX edit by virjoy2001
 ==Video Solution==

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