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

Latest revision as of 16:19, 7 September 2017

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

Solution 1

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 $\frac{1}{2} \cdot \frac{1}{1 - \frac{1}{4}} = \boxed{\textbf{(C)}\; \frac{2}{3}}$ .

Solution 2

Break the problem up into two separate cases: (a) Larry wins on the first throw or (b) Larry wins after the first throw.

a: The probability that Larry wins on the first throw is $\frac{1}{2}$ .

b: The probability that Larry wins after the first throw is half the probability that Julius wins because it only occurs half the time. This probability is $\frac{1}{2}(1-x)$ , where $x$ is the probability that Larry wins.

Therefore, $x = \frac{1}{2} + \frac{1}{2}(1 - x)$ . This equation can be solved for $x$ to find that the probability that Larry wins is $\boxed{\textbf{(C)}\; \frac{2}{3}}$ .

@@ Line 5: / Line 5: @@
 ==Solution==
+==Solution 1==
 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>
-Thus the probability of Larry wining is
+Thus the probability of Larry winning is
 <math>\frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \frac{1}{2^7} + \cdots</math>
-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>.
+This is a geometric series with ratio <math>\frac{1}{2^2}=\frac{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>.
+==Solution 2==
+Break the problem up into two separate cases: (a) Larry wins on the first throw or (b) Larry wins after the first throw.
+a: The probability that Larry wins on the first throw is <math>\frac{1}{2}</math>.
+b: The probability that Larry wins after the first throw is half the probability that Julius wins because it only occurs half the time. This probability is <math>\frac{1}{2}(1-x)</math>, where <math>x</math> is the probability that Larry wins.
+Therefore, <math>x = \frac{1}{2} + \frac{1}{2}(1 - x)</math>. This equation can be solved for <math>x</math> to find that the probability that Larry wins is <math>\boxed{\textbf{(C)}\; \frac{2}{3}}</math>.
 ==See Also==
 {{AMC12 box|year=2015|ab=B|num-a=10|num-b=8}}
 {{MAA Notice}}

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