Difference between revisions of "2015 AMC 12B Problems/Problem 9"
Pi over two (talk | contribs) (→Solution) |
Pi over two (talk | contribs) (→Solution) |
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<math>\frac{1}{2} + \frac{1}{2^3} + \frac{1}{2^5} + \frac{1}{2^7} + \cdots</math> | <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 | + | 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>. |
==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 14:54, 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 
, independently of what has happened before. What is the probability that Larry wins the game?
Solution
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 wining is
This is a geometric series with ratio 
, hence the answer is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
