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Difference between revisions of "2021 AMC 12A Problems/Problem 23"

(→‎Solution 1 (complementary counting))
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==Solution 2 (direct counting and probability states)==
 
==Solution 2 (direct counting and probability states)==
 
We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have <math>1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}</math>, so the answer is <math>\boxed{D}</math>. ~IceWolf10
 
We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have <math>1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}</math>, so the answer is <math>\boxed{D}</math>. ~IceWolf10
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==Solution 3 (Similar to Solution 2, but Find the Numerator and Denominator Separately)==
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<b>Denominator</b>
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There are <math>4^4=256</math> ways to make 4 hops without restrictions.
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<b>Numerator (Casework)</b>
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Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.
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(1) Hop #2 (Hops #3 and #4 have no restrictions)
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The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has <math>4\cdot2\cdot4\cdot4=128</math> ways.
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(2) Hop #3 (Hope #4 has no restriction)
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No matter which direction the first hop takes, the second hop must "wrap around".
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The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has <math>4\cdot1\cdot2\cdot4=32</math> ways.
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(3) Hop #4
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Two sub-cases:
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(3-1) The second hop "wraps around". It follows that the third hop also "wraps around".
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The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has <math>4\cdot1\cdot1\cdot2=8</math> ways.
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(3-2) The second hop backs to the center.
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The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has <math>4\cdot1\cdot4\cdot2=32</math> ways.
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Together, Case (3) has <math>8+32=40</math> ways.
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The numerator is <math>128+32+40=200.</math>
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<b>Probability</b>
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<math>\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.</math>
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This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3
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~MRENTHUSIASM
  
 
== Video Solution by OmegaLearn (Using Probability States) ==
 
== Video Solution by OmegaLearn (Using Probability States) ==

Revision as of 00:05, 12 February 2021

Problem

Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?

$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$

Solution 1 (complementary counting)

We will use complementary counting. First, the frog can go left with probability $\frac14$. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4\cdot \frac14=1$, we will ignore the leading probability.

From the left, she either goes left to another edge ($\frac14$) or back to the center ($\frac14$). Time for some casework.

$\textbf{Case 1:}$ She goes back to the center.

Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac12$. --End case 1

$\textbf{Case 2:}$ She goes to another edge (rightmost).

Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac12$

Subcase 2: She goes to the center. Now any move works.

$\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38$ for this case. --End case 2

She goes back to the center in Case 1 with probability $\frac14$, and to the right edge with probability $\frac14$

So, our answer is $\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}$

But, don't forget complementary counting. So, we get $1-\frac7{32}=\frac{25}{32} \implies \boxed{D}$. ~ firebolt360

Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360

Solution 2 (direct counting and probability states)

We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have $1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}$, so the answer is $\boxed{D}$. ~IceWolf10

Solution 3 (Similar to Solution 2, but Find the Numerator and Denominator Separately)

Denominator

There are $4^4=256$ ways to make 4 hops without restrictions.


Numerator (Casework)

Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.


(1) Hop #2 (Hops #3 and #4 have no restrictions)

The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has $4\cdot2\cdot4\cdot4=128$ ways.


(2) Hop #3 (Hope #4 has no restriction)

No matter which direction the first hop takes, the second hop must "wrap around".

The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has $4\cdot1\cdot2\cdot4=32$ ways.


(3) Hop #4

Two sub-cases:

(3-1) The second hop "wraps around". It follows that the third hop also "wraps around".

The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has $4\cdot1\cdot1\cdot2=8$ ways.

(3-2) The second hop backs to the center.

The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has $4\cdot1\cdot4\cdot2=32$ ways.

Together, Case (3) has $8+32=40$ ways.


The numerator is $128+32+40=200.$


Probability

$\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.$


This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3


~MRENTHUSIASM

Video Solution by OmegaLearn (Using Probability States)

https://youtu.be/V_Sn30N2q50

~ pi_is_3.14

See also

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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
2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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