2017 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 18 Medium

Amelia has a coin that lands on heads with probability \dfrac{1}{3}, and Blaine has a coin that lands on heads with probability \dfrac{2}{5}. 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 \dfrac{p}{q}, where p and q are relatively prime positive integers. What is q- p? (2017 AMC 10A Problem, Question#18)

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    5

Answer:D

Let P be the probability Amelia wins. Note that P= chance she wins on her first turn + chance she gets to her turn again·P, as if she gets to her turn again, she is back where she started with probability of winning P. The chance she wins on her first turn is \dfrac{1}{3}. The chance she makes it to her turn again is a combination of her failing to win the first turn-\dfrac{2}{3} and Blaine failing to win -\dfrac{3}{5}. Multiplying gives us \dfrac{2}{5}. Thus,  P= \dfrac{1}{3}+ \dfrac{2}{5}P. Therefore, p=\dfrac{9}{5}, so the answer is 9-5=(\rm D)4.

Let P be the probability Amelia wins. Note that P= chance she wins on her frst turn + chance she gets to her second turn \cdot\dfrac{1}{3}+ chance she gets to her third turn \cdot\dfrac{1}{3}\ldots. This can be represented by an infinite geometric series. p=\dfrac{\dfrac{1}{3}}{1-\dfrac{2}{3}\cdot\dfrac{3}{6}}=\dfrac{\dfrac{1}{3}}{1-\dfrac{2}{5}}=\dfrac{\dfrac{1}{3}}{\dfrac{3}{5}}=\dfrac{1}{3}\cdot\dfrac{5}{3}=\dfrac{5}{9}. Therefre P= \dfrac{5}{9}, so the answer is 9-5=(\rm D)4.

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