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Difference between revisions of "2023 AMC 10B Problems/Problem 19"
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~Technodoggo
 
~Technodoggo
 +
 +
==Solution 3==
 +
 +
We denote by <math>\left( x, y \right)</math> the frog's initial coordinates.
 +
We denote by <math>k \in \left\{ n, e, s, w \right\}</math> the direction to hop.
 +
We denote by <math>z</math> the hopping distance.
 +
In this analysis, we say that the frog wins if landing outside the square.
 +
 +
We have
 +
<math></math>
 +
\begin{align*}
 +
P \left( win \right)
 +
& = \sum_{k \in \left\{ n, e, s, w \right\}}
 +
P \left( win | k \right) P \left( k \right) \\
 +
& = P \left( win | k = w \right)
 +
\sum_{k \in \left\{ n, e, s, w \right\}}
 +
P \left( k \right) \\
 +
& = P \left( win | k = w \right) \\
 +
& = \int_{y=0}^1 P \left( win | k = w, y \right) dy \\
 +
& = P \left( win | k = w, y = 0 \right) \\
 +
& = P \left( win | k = w, y = 0, x \in [0,1] \right) P \left( x \in [0,1] \right)
 +
+ P \left( win | k = w, y = 0, x \in (1,6] \right) P \left( x \in (1,6] \right) \\
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& = P \left( win | k = w, y = 0, x \in [0,1] \right) \cdot \frac{1}{6}
 +
+ 0 \cdot \frac{5}{6} \\
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& = \frac{1}{6} P \left( win | k = w, y = 0, x \in [0,1] \right) \\
 +
& = \frac{1}{6}
 +
\int_{x=0}^1 \int_{z=x}^1 dz dx \\
 +
& = \frac{1}{6} \cdot \frac{1}{2} \\
 +
& = \boxed{\textbf{(B) <math>\frac{1}{12}</math>}}.
 +
\end{align*}
 +
<math></math>
 +
The second equality above follows from symmetry that <math>P \left( win | k \right)</math> are all the same for all <math>k \in \left\{ n, e, s, w \right\}</math>.
 +
The fifth equality above follows from symmetry that <math>P \left( win | k = w, y \right)</math> are all the same for all <math>y \in \left[ 0, 1 \right]</math>.
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 17:14, 15 November 2023

froggo

Solution 1

WLOG, we assume Sonya jumps $0.5$ units every time, since that is her expected value.

If Sonya is within $0.5$ blocks of an edge, she can jump off the board. Let us examine the region that is at most $0.5$ blocks from exactly one edge.

[asy] import graph; Label f; xaxis(0,6,Ticks(f, 6.0, 0.5)); yaxis(0,6,Ticks(f, 6.0, 0.5)); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); filldraw((0,0.5)--(0.5,0.5)--(0.5,5.5)--(0,5.5)--cycle,gray); filldraw((0.5,0)--(0.5,0.5)--(5.5,0.5)--(5.5,0)--cycle,gray); filldraw((6,0.5)--(5.5,0.5)--(5.5,5.5)--(6,5.5)--cycle,gray); filldraw((0.5,6)--(0.5,5.5)--(5.5,5.5)--(5.5,6)--cycle,gray); [/asy]

If Sonya starts in this region, she has a $\dfrac14$ chance of landing outside (there's exactly one direction she can hop to get out). The total area of this region is $4\cdot0.5\cdot5=10.$ For this region, Sonya has a $\dfrac14$ chance, so we multiply $10$ by $\dfrac14$ to get $2.5.$

If Sonya is in one of the corner squares, she can go two directions to get out, so she has a $\dfrac24=\dfrac12$ chance to get out. The total area is $0.5\cdot0.5\cdot4=1$, so this region yields $\dfrac12\cdot1=\dfrac12.$

Adding the two, we get $3$. This is out of $36$ square units of area, so our answer is thus $\dfrac1{12}.$

~Technodoggo

Solution 2

Since all the actions are independent, we can switch the orders. Let Sonya choose the direction first. And the problem is symmetric, so we consider just one direction. WLOG, let's say she choose $south$. When she first pick the location, she'll have to be within 1 unit of the $x$ axis to have a chance to jump out of the boundary southward. That's $\dfrac{1}{6}$. With in that region, the expected y coordinate would be 0.5 which is 0.5 unit from the boundary (x-axis). Now, the jumping distance required to jump out of the boundary on average has to be greater than 0.5. That's another $\dfrac{1}{2}$. So the final probability is $\dfrac{1}{6}\cdot\dfrac{1}{2} = \dfrac{1}{12}$.

~Technodoggo

Solution 3

We denote by $\left( x, y \right)$ the frog's initial coordinates. We denote by $k \in \left\{ n, e, s, w \right\}$ the direction to hop. We denote by $z$ the hopping distance. In this analysis, we say that the frog wins if landing outside the square.

We have $$ (Error compiling LaTeX. Unknown error_msg) \begin{align*} P \left( win \right) & = \sum_{k \in \left\{ n, e, s, w \right\}} P \left( win | k \right) P \left( k \right) \\ & = P \left( win | k = w \right) \sum_{k \in \left\{ n, e, s, w \right\}} P \left( k \right) \\ & = P \left( win | k = w \right) \\ & = \int_{y=0}^1 P \left( win | k = w, y \right) dy \\ & = P \left( win | k = w, y = 0 \right) \\ & = P \left( win | k = w, y = 0, x \in [0,1] \right) P \left( x \in [0,1] \right) + P \left( win | k = w, y = 0, x \in (1,6] \right) P \left( x \in (1,6] \right) \\ & = P \left( win | k = w, y = 0, x \in [0,1] \right) \cdot \frac{1}{6} + 0 \cdot \frac{5}{6} \\ & = \frac{1}{6} P \left( win | k = w, y = 0, x \in [0,1] \right) \\ & = \frac{1}{6} \int_{x=0}^1 \int_{z=x}^1 dz dx \\ & = \frac{1}{6} \cdot \frac{1}{2} \\ & = \boxed{\textbf{(B) $\frac{1}{12}$}}. \end{align*} $$ (Error compiling LaTeX. Unknown error_msg) The second equality above follows from symmetry that $P \left( win | k \right)$ are all the same for all $k \in \left\{ n, e, s, w \right\}$. The fifth equality above follows from symmetry that $P \left( win | k = w, y \right)$ are all the same for all $y \in \left[ 0, 1 \right]$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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