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

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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>
  
==Solution==
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==Solution 1==
 
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>. 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,
 
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>. 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}P</cmath>
 
<cmath>P = \frac{1}{3} + \frac{2}{5}P</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 2==
 
==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>.
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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
 
Solution by ktong
 +
 +
~minor LaTeX edit by virjoy2001
 +
 +
== Video Solution ==
 +
https://youtu.be/IRyWOZQMTV8?t=4552
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=umr2Aj9ViOA
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2017|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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 +
[[Category:Introductory Probability Problems]]

Latest revision as of 12:37, 10 April 2021

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://youtu.be/IRyWOZQMTV8?t=4552

~ pi_is_3.14

Video Solution

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

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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