Difference between revisions of "2017 AMC 10A Problems/Problem 18"
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Solution by ktong | Solution by ktong | ||
+ | ==Solution 3: Cheap Nonsensical Guessing== | ||
+ | Note that the probabilities as given in the question are 1/3 and 2/5, which altogether consist of the integers {1,2,3,5}. The only integer missing to make this an arithmetic sequence is <math>\boxed{\text{D}\space 4}</math> | ||
==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}} |
Revision as of 14:51, 24 July 2017
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
Solution by ktong
Solution 3: Cheap Nonsensical Guessing
Note that the probabilities as given in the question are 1/3 and 2/5, which altogether consist of the integers {1,2,3,5}. The only integer missing to make this an arithmetic sequence 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 |
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