2016 AMC 12B Problem 19

Attempt Problem 19 of the 2016 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12B solutions, or check the answer key.

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19.

Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times?

18\dfrac18

17\dfrac17

16\dfrac16

14\dfrac14

13\dfrac13

Answer: B
Concepts:geometric distributionindependent events
Difficulty rating: 1910
Solution:

A player's first head comes on flip nn with probability (12)n.\left(\tfrac12\right)^n. All three stopping on the same flip nn has probability ((12)n)3=(18)n.\left(\left(\tfrac12\right)^n\right)^3=\left(\tfrac18\right)^n. Summing over n1,n\ge1, n=1(18)n=1/811/8=17.\displaystyle\sum_{n=1}^\infty\left(\tfrac18\right)^n =\frac{1/8}{1-1/8}=\frac17.

Thus, the correct answer is B.

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