Difference between revisions of "2014 AMC 12B Problems/Problem 22"
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+ | {{duplicate|[[2014 AMC 12B Problems|2014 AMC 12B #22]] and [[2014 AMC 10B Problems|2014 AMC 10B #25]]}} | ||
+ | |||
+ | == Problem == | ||
+ | In a small pond there are eleven lily pads in a row labeled 0 through 10. A frog is sitting on pad 1. When the frog is on pad <math>N</math>, <math>0<N<10</math>, it will jump to pad <math>N-1</math> with probability <math>\frac{N}{10}</math> and to pad <math>N+1</math> with probability <math>1-\frac{N}{10}</math>. Each jump is independent of the previous jumps. If the frog reaches pad 0 it will be eaten by a patiently waiting snake. If the frog reaches pad 10 it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake? | ||
+ | |||
+ | <math>\textbf{(A) }\frac{32}{79}\qquad | ||
+ | \textbf{(B) }\frac{161}{384}\qquad | ||
+ | \textbf{(C) }\frac{63}{146}\qquad | ||
+ | \textbf{(D) }\frac{7}{16}\qquad | ||
+ | \textbf{(E) }\frac{1}{2}\qquad</math> | ||
+ | |||
+ | == Solution 1 == | ||
A long, but straightforward bash: | A long, but straightforward bash: | ||
− | |||
Define <math>P(N)</math> to be the probability that the frog survives starting from pad N. | Define <math>P(N)</math> to be the probability that the frog survives starting from pad N. | ||
− | |||
Then note that by symmetry, <math>P(5) = 1/2</math>, since the probabilities of the frog moving subsequently in either direction from pad 5 are equal. | Then note that by symmetry, <math>P(5) = 1/2</math>, since the probabilities of the frog moving subsequently in either direction from pad 5 are equal. | ||
− | |||
We therefore seek to rewrite <math>P(1)</math> in terms of <math>P(5)</math>, using the fact that | We therefore seek to rewrite <math>P(1)</math> in terms of <math>P(5)</math>, using the fact that | ||
+ | <math>P(N) = \frac {N} {10}P(N - 1) + \frac {10 - N} {10}P(N + 1)</math> | ||
− | + | as said in the problem. | |
+ | Hence <math>P(1) = \frac {1} {10}P(0) + \frac {9} {10}P(2) = \frac {9} {10}P(2)</math> | ||
− | + | <math>\Rightarrow P(2) = \frac {10} {9}P(1)</math> | |
+ | Returning to our original equation: | ||
− | + | <math>P(1) = \frac {9} {10}P(2) = \frac {9} {10}\left(\frac{2} {10}P(1) + \frac{8} {10}P(3)\right)</math> | |
+ | <math>= \frac {9} {50}P(1) + \frac {18} {25}P(3) \Rightarrow P(1) - \frac {9} {50}P(1)</math> | ||
+ | <math>= \frac {18} {25}P(3)</math> | ||
− | <math>\Rightarrow P( | + | <math>\Rightarrow P(3) = \frac {41} {36}P(1)</math> |
− | |||
Returning to our original equation: | Returning to our original equation: | ||
+ | <math>P(1) = \frac {9} {50}P(1) + \frac {18} {25}\left(\frac {3} {10}P(2) + \frac {7} {10}P(4)\right)</math> | ||
− | <math> | + | <math>= \frac {9} {50}P(1) + \frac {27} {125}P(2) + \frac {63} {125}P(4)</math> |
+ | <math>= \frac {9} {50}P(1) + \frac {27} {125}\left(\frac {10} {9}P(1)\right) + \frac {63} {125}\left(\frac {4} {10}P(3) + \frac {6} {10}P(5)\right)</math> | ||
− | + | Cleaing up the coefficients, we have: | |
− | |||
+ | <math>= \frac {21} {50}P(1) + \frac {126} {625}P(3) + \frac {189} {625}P(5)</math> | ||
− | <math>\ | + | <math>= \frac {21} {50}P(1) + \frac {126} {625}\left(\frac {41} {36}P(1)\right) + \frac {189} {625}\left(\frac {1} {2}\right)</math> |
+ | Hence, <math>P(1) = \frac {525} {1250}P(1) + \frac {287} {1250}P(1) + \frac {189} {1250}</math> | ||
− | + | <math>\Rightarrow P(1) - \frac {812} {1250}P(1) = \frac {189} {1250} \Rightarrow P(1) = \frac {189} {438}</math> | |
+ | <math>= \boxed{\frac {63} {146}\, (C)}</math> | ||
− | <math>P(1) = \frac {9} { | + | Or set <math>P(1)=a,P(2)=b,P(3)=c,P(4)=d,P(5)=e=1/2</math>: |
+ | <cmath>a=0.1\emptyset+0.9b,b=0.2a+0.8c,c=0.3b+0.7d,d=0.4c+0.6e</cmath> | ||
+ | <cmath>10a=\emptyset+9b,10b=2a+8c,10c=3b+7d,10d=4c+6e</cmath> | ||
+ | <cmath>\implies b=\frac{10a-\emptyset}{9},c=\frac{5b-a}{4},d=\frac{10c-3b}{7},e=\frac{5d-2c}{3}=1/2</cmath> | ||
+ | <math>b=\frac{10a}{9}</math> | ||
+ | <math>c=\frac{5\left(\frac{10a}{9}\right)-a}{4}=\frac{\frac{50a}{9}-a}{9}=\frac{41a}{36}</math> | ||
− | <math>= \frac { | + | <math>d=\frac{10\left(\frac{41a}{36}\right)-3\left(\frac{30a}{9}\right)}{7}=\frac{\frac{205a}{18}-\frac{10a}{3}}{7}=\frac{145a}{126}</math> |
+ | <math>e=\frac{5\left(\frac{145a}{126}\right)-2\left(\frac{41a}{36}\right)}{3}=\frac{\frac{725a}{126}-\frac{41a}{18}}{3}=\frac{73a}{63}</math> | ||
− | <math>= \frac { | + | Since <math>e=\frac{1}{2}</math>, <math>\frac{73a}{63}=\frac{1}{2}\implies a=\boxed{\textbf{(C) }\frac{63}{146}}</math>. |
+ | == Solution 2 == | ||
+ | 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> chance 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>P_1=\frac{9}{10}\cdot P_2</cmath> | ||
+ | <cmath>P_2=\frac{2}{10}\cdot P_1 + \frac{8}{10}\cdot P_3</cmath> | ||
+ | <cmath>P_3=\frac{3}{10}\cdot P_2 + \frac{7}{10}\cdot P_4</cmath> | ||
+ | <cmath>P_4=\frac{4}{10}\cdot P_3 + \frac{6}{10}\cdot P_5</cmath> | ||
+ | <cmath>P_5=\frac{5}{10}</cmath> | ||
+ | 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>. | ||
− | + | System of equations (alligator112): Start by removing all fractions. From here on out, we will use substitution while avoiding fractions. | |
+ | <cmath>10P_1 = 9P_2</cmath> | ||
+ | <cmath>10P_2 = 2P_1 + 8P_3</cmath> | ||
+ | <cmath>10P_3 = 3P_2 + 7P_4</cmath> | ||
+ | <cmath>10P_4 = 4P_3 + 6P_5 = 4P_3 + 3</cmath> | ||
+ | <cmath>100 P_3 = 30P_2 + 70P_4 = 30P_2 + 7(4P_3 + 3) \rightarrow 72P_3 = 30P_2 + 21</cmath> | ||
+ | <cmath>90P_2 = 18P_1 + 72P_3 = 18P_1 + 30P_2 + 21 \rightarrow 60P_2 = 18P_1 + 21</cmath> | ||
+ | <cmath>200P_1 = 180P_2 = 3(18P_1 + 21) = 54P_1 + 63 \rightarrow P_1 = \frac{63}{146}</cmath> | ||
+ | == Solution 3 == | ||
+ | Assign each lily pad a value, with pad <math>0</math> having value <math>0</math> and pad <math>i > 0</math> having value <math>\sum_{k=0}^{i-1}\frac{1}{\binom{9}{k}}</math>. If we treat the process as a random walk <math>X_0, X_1, \cdots</math> over these values, note that <math>\mathbb{E}[X_{n+1}] = n</math> for all <math>n \ge 0</math>. This is purely by construction, as the expected gain at pad <math>k</math> is <math>\binom{9}{k}\cdot \frac{10-k}{10} - \binom{9}{k-1}\cdot \frac{k}{10} = 0</math>. Therefore the process is a martingale. Since the stopping time has finite expectation and the increments are bounded, the [https://en.wikipedia.org/wiki/Optional_stopping_theorem Optional Stopping Thoerem] applies and the expected final outcome is <math>0</math>. This gives the probability of reaching safety as <math>\left(\sum_{k=0}^{9}\frac{1}{\binom{9}{k}}\right)^{-1} = \frac{63}{146}</math>, which is choice <math>\boxed{(C)}</math>. | ||
− | + | ==Remark (Markov Chain)== | |
+ | We can represent this problem with the following State Transition Diagram of [https://en.wikipedia.org/wiki/Absorbing_Markov_chain Absorbing Markov Chain]. State <math>0</math> and state <math>10</math> are the absorbing states. This problem asks for the [https://en.wikipedia.org/wiki/Absorbing_Markov_chain#Absorbing_probabilities Absorbing Probability] of state <math>10</math>. | ||
− | + | [[File:Markov Chain Frog.png|950px|center]] | |
+ | Let <math>p_{ij} = P(X_{n+1} = j | X_n = i)</math>, the probability that state <math>i</math> transits to state <math>j</math> on the next step. | ||
− | + | Let <math>a_i</math> be the absorbing probability of being absorbed in the absorbing state when starting from transient state <math>i</math>. | |
+ | <math>a_i = \sum_{j} (p_{ij} \cdot a_{j})</math> | ||
− | <math> | + | <math>a_i</math> is the sum of the products of <math>p_{ij}</math> and <math>a_j</math> of all the next state <math>j</math>. |
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=DMdgh2mMiWM | ||
− | + | ==Video Solution by TheBeautyofMath== | |
+ | https://youtu.be/QqeaomXYDsg | ||
+ | ~IceMatrix | ||
− | - | + | == See also == |
+ | {{AMC10 box|year=2014|ab=B|num-b=24|after=Last Question}} | ||
+ | {{AMC12 box|year=2014|ab=B|num-b=21|num-a=23}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:12, 2 August 2024
- The following problem is from both the 2014 AMC 12B #22 and 2014 AMC 10B #25, so both problems redirect to this page.
Contents
Problem
In a small pond there are eleven lily pads in a row labeled 0 through 10. A frog is sitting on pad 1. 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 0 it will be eaten by a patiently waiting snake. If the frog reaches pad 10 it will exit the pond, never to return. What is the probability that the frog will escape without being eaten by the snake?
Solution 1
A long, but straightforward bash:
Define to be the probability that the frog survives starting from pad N.
Then note that by symmetry, , since the probabilities of the frog moving subsequently in either direction from pad 5 are equal.
We therefore seek to rewrite in terms of
, using the fact that
as said in the problem.
Hence
Returning to our original equation:
Returning to our original equation:
Cleaing up the coefficients, we have:
Hence,
Or set :
Since ,
.
Solution 2
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
chance 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:
We want to find
, since the frog starts at pad
. Solving the above system yields
, so the answer is
.
System of equations (alligator112): Start by removing all fractions. From here on out, we will use substitution while avoiding fractions.
Solution 3
Assign each lily pad a value, with pad having value
and pad
having value
. If we treat the process as a random walk
over these values, note that
for all
. This is purely by construction, as the expected gain at pad
is
. Therefore the process is a martingale. Since the stopping time has finite expectation and the increments are bounded, the Optional Stopping Thoerem applies and the expected final outcome is
. This gives the probability of reaching safety as
, which is choice
.
Remark (Markov Chain)
We can represent this problem with the following State Transition Diagram of Absorbing Markov Chain. State and state
are the absorbing states. This problem asks for the Absorbing Probability of state
.
Let , the probability that state
transits to state
on the next step.
Let be the absorbing probability of being absorbed in the absorbing state when starting from transient state
.
is the sum of the products of
and
of all the next state
.
Video Solution
https://www.youtube.com/watch?v=DMdgh2mMiWM
Video Solution by TheBeautyofMath
~IceMatrix
See also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 |
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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.