2015 AMC 12B Problems/Problem 9

Problem

Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is $\tfrac{1}{2}$ , independently of what has happened before. What is the probability that Larry wins the game?

$\textbf{(A)}\; \dfrac{1}{2} \qquad\textbf{(B)}\; \dfrac{3}{5} \qquad\textbf{(C)}\; \dfrac{2}{3} \qquad\textbf{(D)}\; \dfrac{3}{4} \qquad\textbf{(E)}\; \dfrac{4}{5}$

Solution

If Larry wins, he either wins on the first move, or the third move, or the fifth move, etc. Let $W$ represent "player wins", and $L$ represent "player loses". Then the events corresponding to Larry winning are $W, LLW, LLLLW, LLLLLLW, \ldots$

Thus the probability of Larry winning is

$\frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \frac{1}{2^7} + \cdots$

This is a geometric series with ratio $\frac{1}{2^2}=\frac{1}{4}$ , hence the answer is $\frac{1}{2} \cdot \frac{1}{1 - \frac{1}{4}} = \boxed{\textbf{(C)}\; \frac{2}{3}}$ .

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 12 Problems and Solutions

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2015 AMC 12B Problems/Problem 9

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