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Difference between revisions of "2015 AMC 12B Problems/Problem 9"

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==Solution==
 
==Solution==
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If Larry wins, he either wins on the first move, or the third move, or the fifth move, etc. Let <math>W</math> represent "player wins", and <math>L</math> represent "player loses". Then the events corresponding to Larry winning are <math>W, LLW, LLLLW, LLLLLLW, \ldots</math>
  
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Thus the probability of Larry wining is
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<math>\frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \frac{1}{2^7} + \cdots</math>
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This is a geometric series with ratio <math>1/4</math>, hence the answer is <math>\frac{1}{2} \cdot \frac{1}{1 - \frac{1}{4}} = \boxed{\textbf{(C)}\; \frac{2}{3}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=10|num-b=8}}
 
{{AMC12 box|year=2015|ab=B|num-a=10|num-b=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:07, 4 March 2015

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 wining is

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

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

See Also

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

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