Difference between revisions of "2020 AMC 10A Problems/Problem 13"

Revision as of 21:17, 4 November 2024

The following problem is from both the 2020 AMC 12A #11 and 2020 AMC 10A #13, so both problems redirect to this page.

Problem

A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$ , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$ . What is the probability that the sequence of jumps ends on a vertical side of the square?

$\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78$

Solution 3

If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes $\frac{1}{2}$ . Since it starts on $(1,2)$ , there is a $\frac{3}{4}$ chance (up, down, or right) it will reach a diagonal on the first jump and $\frac{1}{4}$ chance (left) it will reach the vertical side. The probablity of landing on a vertical is $\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}$ . - Lingjun

Solution 4 (Complete States)

Let $P_{(x,y)}$ denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at $(x,y)$ . Note that $P_{(1,2)}=P_{(3,2)}$ by reflective symmetry over the line $x=2$ . Similarly, $P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}$ , and $P_{(2,1)}=P_{(2,3)}$ . Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: $P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}$ $P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}$ $P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}$ $P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}$ We have a system of $4$ equations in $4$ variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives $P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)$ $P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}$ Plugging in the third equation into this gives $P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}$ $P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}$ Next, plugging in the second and third equation into the first equation yields $P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)$ $P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}$ Now plugging in (*) into this, we get $P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)$ $P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}$ -mathisawesome2169

Solution 5 (States)

this is basically another version of solution 4; shoutout to mathisawesome2169 :D

First, we note the different places the frog can go at certain locations in the square:

If the frog is at a border vertical point ( $(1,2),(3,2)$ ), it moves with probability $\frac{1}{4}$ to a vertical side of the square, probability $\frac{1}{4}$ to the center of the square, and probability $\frac{1}{2}$ to a corner square.

If the frog is at a border horizontal point ( $(2,1),(2,3)$ ), it moves with probability $\frac{1}{4}$ to a horizontal side of the square, probability $\frac{1}{4}$ to the center of the square, and probability $\frac{1}{2}$ to a corner square.

If the frog is at a center square ( $(2,2)$ ), it moves with probability $\frac{1}{2}$ to a border horizontal point and probability $\frac{1}{2}$ to a border vertical point.

If the frog is at a corner ( $(1,1),(1,3),(3,3),(3,1)$ ), it moves with probability $\frac{1}{4}$ to a vertical side of the square, probability $\frac{1}{4}$ to a horizontal side, probability $\frac{1}{4}$ to a border horizontal point, and probability $\frac{1}{4}$ to a border vertical point.

Now, let $x$ denote the probability of the frog reaching a vertical side when it is at a border vertical point. Similarly, let $y$ denote the probability of the frog reaching a vertical side when it is at a border horizontal point. Now, the probability of the frog reaching a vertical side of the square at any location inside the square can be expressed in terms of $x$ and $y$ .

First, the two easier ones: $P_{center}=\frac{1}{2}x+\frac{1}{2}y$ , and $P_{corner}=\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}y$ . Now, we can write $x$ and $y$ in terms of $x$ and $y$ , allowing us to solve a system of two variables: $x=\frac{1}{4}+\frac{1}{4}P_{center}+\frac{1}{2}P_{corner}=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}y\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}y\right)$ and $y=\frac{1}{4}P_{center}+\frac{1}{2}P_{corner}=\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}y\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}y\right).$ From these two equations, it is apparent that $y=x-\frac{1}{4}$ . We can then substitute this value for $y$ back into any of the two equations above to get $x=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}\left(x-\frac{1}{4}\right)\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}\left(x-\frac{1}{4}\right)\right).$ Although this certainly looks intimidating, we can expand the parentheses and multiply both sides by 16 to eliminate the fractions, which upon simplification yields the equation $16x=5+8x,$ giving us the desired probability $x=\frac{5}{8}$ . The answer is then $\boxed{\textbf{(B) }\frac{5}{8}}$ .

- curiousmind888 & TGSN

Video Solutions

Video Solution 1

IceMatrix's Solution (Starts at 4:40)

Video Solution 2 (Simple & Quick)

https://youtu.be/qNaN0BlIsw0

Video Solution 3

On The Spot STEM

https://youtu.be/xGs7BjQbGYU

Video Solution 4

https://youtu.be/0m4lbXSUV1I

~savannahsolver

Video Solution 5

https://youtu.be/IRyWOZQMTV8?t=5173

~ pi_is_3.14

Video Solution 6

https://www.youtube.com/watch?v=R220vbM_my8

~ amritvignesh0719062.0

Video Solution 7

https://www.youtube.com/watch?v=TvYoiU_zct8

~ mathgenius2012

@@ Line 7: / Line 7: @@
 <math>\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78</math>
-==Solution 1==
+==Solution 3==
 If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes <math>\frac{1}{2}</math>. Since it starts on <math>(1,2)</math>, there is a <math>\frac{3}{4}</math> chance (up, down, or right) it will reach a diagonal on the first jump and <math>\frac{1}{4}</math> chance (left) it will reach the vertical side. The probablity of landing on a vertical is <math>\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}</math>.
 - Lingjun
-==Solution 2 (Complete States)==
+==Solution 4 (Complete States)==
 Let <math>P_{(x,y)}</math> denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at <math>(x,y)</math>. Note that <math>P_{(1,2)}=P_{(3,2)}</math> by reflective symmetry over the line <math>x=2</math>. Similarly, <math>P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}</math>, and <math>P_{(2,1)}=P_{(2,3)}</math>.
 Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point:
@@ Line 32: / Line 32: @@
 -mathisawesome2169
-==Solution 3 (States)==
+==Solution 5 (States)==
 this is basically another version of solution 4; shoutout to mathisawesome2169 :D

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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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