Difference between revisions of "2014 AMC 10B Problems/Problem 25"
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==Solution== | ==Solution== | ||
| − | Notice that the probabilities are symmetrical around the fifth lily pad. If the frog is on the fifth lily pad, there is a <math>\frac{1}{2}</math> chance that it escapes and a <math>\frac{1}{2}</math> that it gets eaten. Now, let <math> | + | Notice that the probabilities are symmetrical around the fifth lily pad. If the frog is on the fifth lily pad, there is a <math>\frac{1}{2}</math> chance that it escapes and a <math>\frac{1}{2}</math> that it gets eaten. Now, let <math>P_k</math> represent the probability that the frog escapes if it is currently on pad <math>k</math>. We get the following system of <math>5</math> equations: |
| − | <cmath> | + | <cmath>P_1=\frac{9}{10}\cdot P_2</cmath> |
| − | <cmath> | + | <cmath>P_2=\frac{1}{5}\cdot P_1 + \frac{4}{5}\cdot P_3</cmath> |
| − | <cmath> | + | <cmath>P_3=\frac{3}{10}\cdot P_2 + \frac{7}{10}\cdot P_4</cmath> |
| − | <cmath> | + | <cmath>P_4=\frac{2}{5}\cdot P_3 + \frac{3}{5}\cdot P_5</cmath> |
| − | <cmath> | + | <cmath>P_5=\frac{1}{2}</cmath> |
| − | We want to find <math> | + | We want to find <math>P_1</math>, since the frog starts at pad <math>1</math>. Solving the above system yields <math>P_1=\frac{63}{146}</math>, so the answer is <math>\boxed{(C)}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=24|after=Last Problem}} | {{AMC10 box|year=2014|ab=B|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:16, 10 July 2015
Problem
In a small pond there are eleven lily pads in a row labeled 
through 
. A frog is sitting on pad 
. When the frog is on pad 
, 
, it will jump to pad 
with probability 
and to pad 
with probability 
. Each jump is independent of the previous jumps. If the frog reaches pad it will be eaten by a patiently waiting snake. If the frog reaches pad it will exit the pond, never to return. What is the probability that the frog will escape being eaten by the snake?
Solution
Notice that the probabilities are symmetrical around the fifth lily pad. If the frog is on the fifth lily pad, there is a 
chance that it escapes and a that it gets eaten. Now, let 
represent the probability that the frog escapes if it is currently on pad 
. We get the following system of 
equations:
![\[P_1=\frac{9}{10}\cdot P_2\]](http://latex.artofproblemsolving.com/9/c/2/9c2e4c28a45884b84086cb6eb0f167d4d3e69db5.png)
![\[P_2=\frac{1}{5}\cdot P_1 + \frac{4}{5}\cdot P_3\]](http://latex.artofproblemsolving.com/f/2/e/f2e26b704390ec36cae35bcb2fe1cca2a4a3aa66.png)
![\[P_3=\frac{3}{10}\cdot P_2 + \frac{7}{10}\cdot P_4\]](http://latex.artofproblemsolving.com/e/6/5/e65e1fbaa78bf606934653934fefca3a56818f16.png)
![\[P_4=\frac{2}{5}\cdot P_3 + \frac{3}{5}\cdot P_5\]](http://latex.artofproblemsolving.com/1/5/5/1550ad960fc78d98eecd123e4faf0eda93af5d20.png)
![\[P_5=\frac{1}{2}\]](http://latex.artofproblemsolving.com/2/2/2/222309b35543edb2366b2013d4af855612bc9693.png)
We want to find 
, since the frog starts at pad . Solving the above system yields 
, so the answer is 
.
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 10 Problems and Solutions | ||
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