Difference between revisions of "2017 AMC 10A Problems/Problem 18"
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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 | + | 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> | ||
Revision as of 22:12, 8 February 2017
Contents
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
Let be the probability Amelia wins. Note that
, as 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
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.