2017 AMC 10A Problems/Problem 18
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 | |
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