Difference between revisions of "2019 AIME I Problems/Problem 5"

Revision as of 12:45, 12 February 2021

Problem 5

A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ , $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\frac{1}{3}$ , independently of its previous moves. The probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$ , where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$ . Find $m + n$ .

Solution 1

One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as $P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)$ for $x,y \geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero. We then recursively find $P(4,4) = \frac{245}{2187}$ so the answer is $245 + 7 = \boxed{252}$ .

If this algebra seems intimidating, you can watch a nice pictorial explanation of this by On The Spot Stem. https://www.youtube.com/watch?v=XBRuy3_TM9w

Solution 2

Obviously, the only way to reach (0,0) is to get to (1,1) and then have a $\frac{1}{3}$ chance to get to (0,0). Let x denote a move left 1 unit, y denote a move down 1 unit, and z denote a move left and down one unit each. The possible cases for these moves are $(x,y,z)=(0,0,3),(1,1,2),(2,2,1)$ and $(3,3,0)$ . This gives a probability of $1 \cdot \frac{1}{27} + \frac{4!}{2!} \cdot \frac{1}{81} + \frac{5!}{2! \cdot 2!} \cdot \frac{1}{243} +\frac{6!}{3! \cdot 3!} \cdot \frac{1}{729}=\frac{245}{729}$ to get to $(1,1)$ . The probability of reaching $(0,0)$ is $\frac{245}{3^7}$ . This gives $245+7=\boxed{252}$ .

Solution 3

Since the particle stops at one of the axes, we know that the particle most pass through $(1,1)$ . Thus, it suffices to consider the probability our particle will reach $(1,1)$ . Denote a move to the left, down, diagonally, as X,Y,Z, respectively. Then the only ways to get to $(1,1)$ from $(4,4)$ are the following:

$(1) 0X 0Y 3Z \$ (Error compiling LaTeX. Unknown error_msg)(2) 1X 1Y 2Z \$ $(3) 2X 2Y 1Z \$ (Error compiling LaTeX. Unknown error_msg)(4) 3X 3Y 0Z \$

The probability of (1) is $\frac{1}{3^3}$ . The probability of (2) is $\frac{\frac{4!}{2!}}{3^4} = \frac{12}{3^4}$ . The probability of (3) is $\frac{\frac{5!}{2!2!}}{3^5} = \frac{30}{3^5}$ . The probability of (4) is $\frac{\frac{6!}{3!3!}}{3^6} = \frac{20}{3^6}$ . Adding all of these together, we obtain a total probability of $\frac{245}{3^6}$ that our particle will hit $(1,1)$ . Trivially, there is a $\frac{1}{3}$ chance our particle will hit $(0,0)$ from $(1,1)$ . So our final probability will be $\frac{245}{3^6} \cdot \frac{1}{3} = \frac{245}{3^7} \implies m = 245, n = 7 \implies \boxed{252}$

~NotSoTrivial

Video Solution

Unique solution: https://youtu.be/I-8xZGhoDUY

~Shreyas S

@@ Line 20: / Line 20: @@
 Since the particle stops at one of the axes, we know that the particle most pass through <math>(1,1)</math>. Thus, it suffices to consider the probability our particle will reach <math>(1,1)</math>. Denote a move to the left, down, diagonally, as X,Y,Z, respectively. Then the only ways to get to <math>(1,1)</math> from <math>(4,4)</math> are the following:
-(1) 0X 0Y 3Z \\
+<math>(1) 0X 0Y 3Z \\$
-(2) 1X 1Y 2Z \\
+</math>(2) 1X 1Y 2Z \\$
-(3) 2X 2Y 1Z \\
+<math>(3) 2X 2Y 1Z \\$
-(4) 3X 3Y 0Z \\
+</math>(4) 3X 3Y 0Z \\$
-The probability of (1) is \frac{1}{3^3}. The probability of (2) is <math>\frac{\frac{4!}{2!}}{3^4} = \frac{12}{3^4}</math>. The probability of (3) is <math>\frac{\frac{5!}{2!2!}}{3^5} = \frac{30}{3^5}</math>. The probability of (4) is <math>\frac{\frac{6!}{3!3!}}{3^6} = \frac{20}{3^6}</math>. Adding all of these together, we obtain a total probability of <math>\frac{245}{3^6}</math> that our particle will hit (1,1). Trivially, there is a <math>\frac{1}{3}</math> chance our particle will hit (0,0) from (1,1). So our final probability will be <math>\frac{245}{3^6} \cdot \frac{1}{3} = \frac{245}{3^7} \implies m = 245, n = 7 \implies \boxed{252}</math>
+The probability of (1) is <math>\frac{1}{3^3}</math>. The probability of (2) is <math>\frac{\frac{4!}{2!}}{3^4} = \frac{12}{3^4}</math>. The probability of (3) is <math>\frac{\frac{5!}{2!2!}}{3^5} = \frac{30}{3^5}</math>. The probability of (4) is <math>\frac{\frac{6!}{3!3!}}{3^6} = \frac{20}{3^6}</math>. Adding all of these together, we obtain a total probability of <math>\frac{245}{3^6}</math> that our particle will hit <math>(1,1)</math>. Trivially, there is a <math>\frac{1}{3}</math> chance our particle will hit <math>(0,0)</math> from <math>(1,1)</math>. So our final probability will be <math>\frac{245}{3^6} \cdot \frac{1}{3} = \frac{245}{3^7} \implies m = 245, n = 7 \implies \boxed{252}</math>
 ~NotSoTrivial

2019 AIME I (Problems • Answer Key • Resources)
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