Difference between revisions of "2015 AMC 12B Problems/Problem 9"
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==Solution== | ==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> | 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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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>. | 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 \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}}. | ||
==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 16:16, 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 , independently of what has happened before. What is the probability that Larry wins the game?
Solution
Solution 1
If Larry wins, he either wins on the first move, or the third move, or the fifth move, etc. Let represent "player wins", and
represent "player loses". Then the events corresponding to Larry winning are
Thus the probability of Larry winning is
This is a geometric series with ratio , hence the answer is
.
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}}.
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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