Difference between revisions of "2017 AMC 10A Problems/Problem 18"
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==Solution 1== | ==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>, | + | 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>, since 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> | + | 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> | ||
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~minor LaTeX edit by virjoy2001 | ~minor LaTeX edit by virjoy2001 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/IRyWOZQMTV8?t=4552 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
− | [[Category: | + | [[Category:Introductory Probability Problems]] |
Latest revision as of 01:28, 24 July 2021
Problem
Amelia has a coin that lands heads with probability , and Blaine has a coin that lands on heads with probability . 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 , where and are relatively prime positive integers. What is ?
Solution 1
Let be the probability Amelia wins. Note that , since if she gets to her turn again, she is back where she started with probability of winning . The chance she wins on her first turn is . The chance she makes it to her turn again is a combination of her failing to win the first turn - and Blaine failing to win - . Multiplying gives us . Thus, Therefore, , so the answer is .
Solution 2
Let be the probability Amelia wins. Note that This can be represented by an infinite geometric series: Therefore, , so the answer is
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 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.