Difference between revisions of "2017 AMC 10A Problems/Problem 18"
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Revision as of 12:37, 10 April 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
, 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
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 |
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