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Difference between revisions of "2022 AMC 10B Problems/Problem 23"

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(→‎Remark (Rigorous Calculus))
 
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==Solution==
+
{{duplicate|[[2022 AMC 10B Problems/Problem 23|2022 AMC 10B #23]] and [[2022 AMC 12B Problems/Problem 22|2022 AMC 12B #22]]}}
  
−
We use the following lemma to solve this problem.
+
==Problem==
 +
Ant Amelia starts on the number line at <math>0</math> and crawls in the following manner. For <math>n=1,2,3,</math> Amelia chooses a time duration <math>t_n</math> and an increment <math>x_n</math> independently and uniformly at random from the interval <math>(0,1).</math> During the <math>n</math>th step of the process, Amelia moves <math>x_n</math> units in the positive direction, using up <math>t_n</math> minutes. If the total elapsed time has exceeded <math>1</math> minute during the <math>n</math>th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most <math>3</math> steps in all. What is the probability that Amelia’s position when she stops will be greater than <math>1</math>?
  
−
---------------------------------------
+
<math>\textbf{(A) }\frac{1}{3} \qquad \textbf{(B) }\frac{1}{2} \qquad \textbf{(C) }\frac{2}{3} \qquad \textbf{(D) }\frac{3}{4} \qquad \textbf{(E) }\frac{5}{6}</math>
−
Let <math>y_1, y_2, \cdots, y_n</math> be independent random variables that are uniformly distributed on <math>(0,1)</math>. Then for <math>n = 2</math>,
 
−
<cmath>
 
−
\[
 
−
\Bbb P \left( y_1 + y_2 \leq 1 \right) = \frac{1}{2} .
 
−
\]
 
−
</cmath>
 
  
−
For <math>n = 3</math>,
+
==Solution 1 (Geometric Probability)==
−
<cmath>
 
−
\[
 
−
\Bbb P \left( y_1 + y_2 + y_3 \leq 1 \right) = \frac{1}{6} .
 
−
\]
 
−
</cmath>
 
−
---------------------------------------
 
  
−
Now, we solve this problem.
+
Let <math>x</math> and <math>y</math> be random variables that are independently and uniformly distributed in the interval <math>(0,1).</math> Note that <cmath>P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,</cmath> as shown below:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(200);
  
−
We denote by <math>\tau</math> the last step Amelia moves. Thus, <math>\tau \in \left\{ 2, 3 \right\}</math>.
+
real xMin = -0.25;
−
We have
+
real xMax = 1.25;
 +
real yMin = -0.25;
 +
real yMax = 1.25;
  
−
<cmath>
+
//Draws the horizontal ticks
−
\begin{align*}
+
void horizontalTicks()
−
P \left( \sum_{n=1}^\tau x_n > 1 \right)
+
{
−
& = P \left( x_1 + x_2 > 1 | t_1 + t_2 > 1 \right)
+
  for (real i = 1; i < yMax; ++i)
−
P \left( t_1 + t_2 > 1 \right) \\
+
  {
−
& \hspace{1cm} + P \left( x_1 + x_2 + x_3 > 1 | t_1 + t_2 \leq 1 \right)
+
    draw((-1/32,i)--(1/32,i), black+linewidth(1));
−
P \left( t_1 + t_2 \leq 1 \right) \\
+
  }
−
& = P \left( x_1 + x_2 > 1 \right)
+
}
−
P \left( t_1 + t_2 > 1 \right)
 
−
+ P \left( x_1 + x_2 + x_3 > 1 \right)
 
−
P \left( t_1 + t_2 \leq 1 \right) \\
 
−
& = \left( 1 - \frac{1}{2} \right)\left( 1 - \frac{1}{2} \right)
 
−
+ \left( 1 - \frac{1}{6} \right) \frac{1}{2} \\
 
−
\end{align*}
 
−
</cmath>
 
  
−
<math></math>
+
//Draws the vertical ticks
−
\begin{align*}
+
void verticalTicks()
 +
{
 +
  for (real i = 1; i < xMax; ++i)
 +
  {
 +
    draw((i,-1/32)--(i,1/32), black+linewidth(1));
 +
  }
 +
}
  
 +
horizontalTicks();
 +
verticalTicks();
 +
label("$0$",(0,0),2*SW);
 +
label("$1$",(1,0),2*S);
 +
label("$1$",(0,1),2*W);
 +
fill((0,0)--(1,0)--(0,1)--cycle,yellow);
 +
draw((0,1)--(1,1)^^(1,0)--(1,1),dashed);
 +
draw((0,1)--(1,0));
 +
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(8));
 +
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(8));
 +
label("$x$",(xMax,0),(2,0));
 +
label("$y$",(0,yMax),(0,2));
 +
</asy>
 +
Let <math>x,y,</math> and <math>z</math> be random variables that are independently and uniformly distributed in the interval <math>(0,1).</math> Note that <cmath>P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,</cmath> as shown below:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(200);
 +
import graph3;
 +
import solids;
  
 +
currentprojection=orthographic((0.3,0.1,0.1));
  
−
& = \boxed{\textbf{(C) <math>\frac{2}{3}</math>}} ,
+
draw(surface((1,0,0)--(0,1,0)--(0,0,1)--cycle),yellow);
−
\end{align*}
+
draw(surface((1,0,0)--(0,1,0)--(0,0,0)--cycle),yellow);
−
<math></math>
+
draw(surface((1,0,0)--(0,0,1)--(0,0,0)--cycle),yellow);
 +
draw(surface((0,1,0)--(0,0,1)--(0,0,0)--cycle),yellow);
  
−
where the second equation follows from the property that <math>\left\{ x_n \right\}</math> and <math>\left\{ t_n \right\}</math> are independent sequences, the third equality follows from the lemma above.
+
draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--cycle,dashed);
 +
draw((0,1,0)--(1,1,0)--(1,0,0),dashed);
 +
draw((0,1,1)--(0,1,0)^^(1,1,1)--(1,1,0)^^(1,0,1)--(1,0,0),dashed);
 +
draw((-0.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10));
 +
draw((0,-0.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10));
 +
draw((0,0,-0.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10));
 +
draw((-0.1,0,1)--(0.1,0,1),linewidth(1));
 +
draw((0,1,-0.1)--(0,1,0.1),linewidth(1));
 +
draw((1,-0.1,0)--(1,0.1,0),linewidth(1));
 +
label("$x$",(1.5,0,0),4*dir((1.5,0,0)));
 +
label("$y$",(0,1.5,0),2*dir((0,1.5,0)));
 +
label("$z$",(0,0,1.5),2*dir((0,0,1.5)));
 +
label("$0$",(0,0,0),2*dir((0,0.5,-0.5)));
 +
label("$1$",(1,0,0),4*dir((0,-1,0)));
 +
label("$1$",(0,1,0),4*dir((0,0,-1)));
 +
label("$1$",(0,0,1),5*dir((-1,0,0)));
 +
draw((1,0,0)--(0,1,0)--(0,0,1)--cycle);
 +
</asy>
 +
We have two cases:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>Amelia takes exactly <math>2</math> steps.</li><p>
 +
We need <math>x_1+x_2>1</math> and <math>t_1+t_2>1.</math> So, the probability is <cmath>P(x_1+x_2>1)\cdot P(t_1+t_2>1)=\left(1-\frac12\right)\cdot\left(1-\frac12\right)=\frac14.</cmath>
 +
  <li>Amelia takes exactly <math>3</math> steps.</li><p>
 +
We need <math>x_1+x_2+x_3>1</math> and <math>t_1+t_2\leq1.</math> So, the probability is <cmath>P(x_1+x_2+x_3>1)\cdot P(t_1+t_2\leq1)=\left(1-\frac16\right)\cdot\frac12=\frac{5}{12}.</cmath>
 +
</ol>
 +
Together, the answer is <math>\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}.</math>
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
~MRENTHUSIASM
 +
 +
==Solution 2 (Generalization and Induction)==
 +
 +
We can in fact find the probability that any number of randomly distributed numbers on the interval <math>[0, 1]</math> sum to more than <math>1</math> using geometric probability, as shown in the video below.
 +
 +
If we graph the points that satisfy <math>x + y < 1</math>, <math>0 < x, y < 1</math>, we get the triangle with points <math>(0, 0)</math>, <math>(1, 0)</math>, and <math>(0, 1)</math>. If we graph the points that satisfy <math>x + y + z < 1</math>, <math>0 < x, y, z < 1</math>, we get the tetrahedron with points <math>(0, 0, 0)</math>, <math>(1, 0, 0)</math>, <math>(0, 1, 0)</math>, and <math>(0, 0, 1)</math>.
 +
 +
Of course, the probability of either of these cases happening is simply the area/volume of the points we graphed divided by the total area of the graph, which is always <math>1</math> (this would be much simpler than my calculus proof above).
 +
 +
Thus, we can now solve for the probability that the sum is less than one for <math>n</math> numbers using induction.
 +
 +
<math>\textbf{Claim:}</math> The probability that the sum is less than one is <math>\frac{1}{n!}</math>.
 +
 +
<math>\textbf{Base Case:}</math> For just <math>1</math> number, the probability is <math>1</math>.
 +
 +
<math>\textbf{Induction step:}</math> Suppose that the probability for <math>n</math> numbers is <math>\frac{1}{n!}</math>. We will prove that the probability for <math>n+1</math> numbers is <math>\frac{1}{(n+1)!}</math>. To prove this, we consider that the area of an <math>n+1</math>-dimensional tetrahedron is simply the area/volume of the base times the height divided by <math>n+1</math>.
 +
 +
Of course, the area of the base is <math>\frac{1}{n!}</math>, and the height is <math>1</math>, and thus, we obtain <math>\frac{1}{n! \cdot (n+1)} = \frac{1}{(n+1)!}</math> as our volume (this may be hard to visualize for higher dimensions). The induction step is complete.
 +
 +
The probability of the sum being less than <math>1</math> is <math>\frac{1}{n!}</math>, and the probability of the sum being more than <math>1</math> is <math>\frac{n!-1}{n!}</math>. This trivializes the problem. The answer is <cmath>\frac{1}{2} \cdot \frac{2! - 1}{2!} + \frac{1}{2} \cdot \frac{3! - 1}{3!} = \boxed{\textbf{(C) }\frac{2}{3}}.</cmath>
 +
 +
~mathboy100
 +
 +
==Solution 3 (Observations)==
 +
There are two cases: Amelia takes two steps or three steps.
 +
 +
The former case has a probability of <math>\frac{1}{2}</math>, as stated in Solution 1, and thus the latter also has a probability of <math>\frac{1}{2}</math>.
 +
 +
The probability that Amelia passes <math>1</math> after two steps is also <math>\frac{1}{2}</math>, as it is symmetric to the probability above.
 +
 +
Thus, if the probability that Amelia passes <math>1</math> after three steps is <math>x</math>, our total probability is <math>\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot x</math>. We know that <math>0 < x < 1</math>, and it is relatively obvious that <math>x > \frac{1}{2}</math> (because the probability that <math>x > \frac{3}{2}</math> is <math>\frac{1}{2}</math>). This means that our total probability is between <math>\frac{1}{2}</math> and <math>\frac{3}{4}</math>, non-inclusive, so the only answer choice that fits is <math>\boxed{\textbf{(C) }\frac{2}{3}}</math>.
 +
 +
~mathboy100
 +
 +
==Remark (Calculus)==
 +
It is not immediately clear why three random numbers between <math>0</math> and <math>1</math> have a probability of <math>\frac{5}{6}</math> of summing to more than <math>1</math>. Here is a proof:
 +
 +
Let us start by finding the probability that two random numbers between <math>0</math> and <math>1</math> have a sum of more than <math>x</math>, where <math>0 \leq x \leq 1</math>.
 +
 +
Suppose that our two numbers are <math>y</math> and <math>z</math>. Then, the probability that <math>y > x</math> (which means that <math>y + z > x</math>) is <math>1 - x</math>, and the probability that <math>y < x</math> is <math>x</math>.
 +
 +
If <math>y < x</math>, the probability that <math>y + z > x</math> is <math>1 - x + y</math>. This is because the probability that <math>y + z < x</math> is equal to the probability that <math>z < x - y</math>, which is <math>x - y</math>, so our total probability is <math>1 - (x - y) = 1 - x + y</math>.
 +
 +
Let us now find the average of the probability that <math>y + z > x</math> when <math>y < x</math>. Since <math>y</math> is a random number between <math>0</math> and <math>x</math>, its average is <math>\frac{x}{2}</math>. Thus, our average is <math>1 - x + \frac{1}{2} = 1 - \frac{x}{2}</math>.
 +
 +
Hence, our total probability is equal to
 +
<cmath>1(1-x) + \left(1 - \frac{x}{2}\right)(x) = 1 - \frac{1}{2}x^2.</cmath>
 +
Now, let us find the probability that three numbers uniformly distributed between <math>0</math> and <math>1</math> sum to more than <math>1</math>.
 +
 +
Let our three numbers be <math>a</math>, <math>b</math>, and <math>c</math>. Then, the probability that <math>a + b + c > 1</math> is equal to the probability that <math>b + c</math> is greater than <math>1 - a</math>, which is equal to <math>1 - \frac{1}{2}(1 - a)^2</math>.
 +
 +
To find the total probability, we must average over all values of <math>a</math>. This average is simply equal to the area under the curve <math>1 - \frac{1}{2}(1-x)^2</math> from <math>0</math> to <math>1</math>, all divided by <math>1</math>. We can compute this value using integrals:
 +
<cmath>\begin{align*}
 +
\frac{\int_0^1 \! 1 - \frac{1}{2}(1 - x)^2 \mathrm{d}x}{1} &= \int_0^1 \! 1 - \frac{1}{2}(1 - x)^2 \mathrm{d}x \\
 +
&= 1 - \frac{1}{2}\int_0^1 \! (1 - x)^2 \mathrm{d}x \\
 +
&= 1 - \frac{1}{2}\int_0^1 \! x^2 \mathrm{d}x \\
 +
&= 1 - \frac{1}{2}\left(\frac{1}{3}\right) \\
 +
&= \frac{5}{6}.
 +
\end{align*}</cmath>
 +
For those who don't know calculus, <math>\int_m^n \! f(x) \mathrm{d}x</math> is the area under the curve <math>f(x)</math> from <math>m</math> to <math>n</math>.
 +
 +
~mathboy100
 +
 +
==Remark (Rigorous Calculus)==
 +
 +
In the language of probability, we have three random variables <math>X</math>, <math>Y</math>, <math>Z</math>, each independent and uniformly distributed over the interval <math>(0, 1)</math>. We are interested in the probability that <math>X+Y+Z>1</math>, or <math>P(X+Y+Z>1)</math>.
 +
 +
It follows from the definition of the joint probability density function <math>f(x, y, z)</math> that (we assume that <math>0<x, y, z<1</math>):
 +
\begin{align}
 +
P(X+Y+Z>1)=\iiint_{x+y+z>1} f(x,y,z) \,dx\,dy\,dz.
 +
\end{align}
 +
Since <math>X, Y, Z</math> are independent, we have <math>f(x, y, z)=f_X(x)*f_Y(y)*f_Z(z)</math>, where <math>f_X(x), f_Y(y), f_Z(z)</math> represent the probability densities of <math>X, Y, Z</math> respectively. Recall that <math>X, Y, Z</math> are uniformly distributed over the interval <math>(0, 1)</math>. Hence, <math>f_X(x)=f_Y(y)=f_Z(z)=1</math>.
 +
 +
It remains to evaluate the following triple integral
 +
\begin{align}
 +
\iiint_{x+y+z>1} \,dx\,dy\,dz=\frac{5}{6}.
 +
\end{align}
 +
Therefore, <math>P(X+Y+Z>1)=\frac{5}{6}</math>.
 +
 +
~tsun26
 +
 +
== Video Solution by OmegaLearn Using Geometric Probability ==
 +
https://youtu.be/-AqhcVX8mTw
 +
 +
~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==
 +
 +
https://youtu.be/WsA94SmsF5o
 +
 +
~ThePuzzlr
  
 
https://youtu.be/qOxnx_c9kVo
 
https://youtu.be/qOxnx_c9kVo
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Video Solution by The Power of Logic (#22 and #23)==
 +
https://youtu.be/rZaJSTbs7jY
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/KRkNnlszdEg
 +
 +
~Interstigation
 +
 +
==See Also==
 +
{{AMC10 box|year=2022|ab=B|num-b=22|num-a=24}}
 +
{{AMC12 box|year=2022|ab=B|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Latest revision as of 21:09, 4 November 2024

The following problem is from both the 2022 AMC 10B #23 and 2022 AMC 12B #22, so both problems redirect to this page.

Problem

Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n=1,2,3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0,1).$ During the $n$th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1$?

$\textbf{(A) }\frac{1}{3} \qquad \textbf{(B) }\frac{1}{2} \qquad \textbf{(C) }\frac{2}{3} \qquad \textbf{(D) }\frac{3}{4} \qquad \textbf{(E) }\frac{5}{6}$

Solution 1 (Geometric Probability)

Let $x$ and $y$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,\] as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); real xMin = -0.25; real xMax = 1.25; real yMin = -0.25; real yMax = 1.25; //Draws the horizontal ticks void horizontalTicks() { for (real i = 1; i < yMax; ++i) { draw((-1/32,i)--(1/32,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = 1; i < xMax; ++i) { draw((i,-1/32)--(i,1/32), black+linewidth(1)); } } horizontalTicks(); verticalTicks(); label("$0$",(0,0),2*SW); label("$1$",(1,0),2*S); label("$1$",(0,1),2*W); fill((0,0)--(1,0)--(0,1)--cycle,yellow); draw((0,1)--(1,1)^^(1,0)--(1,1),dashed); draw((0,1)--(1,0)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(8)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(8)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); [/asy] Let $x,y,$ and $z$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,\] as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); import graph3; import solids; currentprojection=orthographic((0.3,0.1,0.1)); draw(surface((1,0,0)--(0,1,0)--(0,0,1)--cycle),yellow); draw(surface((1,0,0)--(0,1,0)--(0,0,0)--cycle),yellow); draw(surface((1,0,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw(surface((0,1,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--cycle,dashed); draw((0,1,0)--(1,1,0)--(1,0,0),dashed); draw((0,1,1)--(0,1,0)^^(1,1,1)--(1,1,0)^^(1,0,1)--(1,0,0),dashed); draw((-0.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-0.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-0.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); draw((-0.1,0,1)--(0.1,0,1),linewidth(1)); draw((0,1,-0.1)--(0,1,0.1),linewidth(1)); draw((1,-0.1,0)--(1,0.1,0),linewidth(1)); label("$x$",(1.5,0,0),4*dir((1.5,0,0))); label("$y$",(0,1.5,0),2*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); label("$0$",(0,0,0),2*dir((0,0.5,-0.5))); label("$1$",(1,0,0),4*dir((0,-1,0))); label("$1$",(0,1,0),4*dir((0,0,-1))); label("$1$",(0,0,1),5*dir((-1,0,0))); draw((1,0,0)--(0,1,0)--(0,0,1)--cycle); [/asy] We have two cases:

  1. Amelia takes exactly $2$ steps.

    2. We need $x_1+x_2>1$ and $t_1+t_2>1.$ So, the probability is \[P(x_1+x_2>1)\cdot P(t_1+t_2>1)=\left(1-\frac12\right)\cdot\left(1-\frac12\right)=\frac14.\]

  2. Amelia takes exactly $3$ steps.

    3. We need $x_1+x_2+x_3>1$ and $t_1+t_2\leq1.$ So, the probability is \[P(x_1+x_2+x_3>1)\cdot P(t_1+t_2\leq1)=\left(1-\frac16\right)\cdot\frac12=\frac{5}{12}.\]

Together, the answer is $\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}.$

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

~MRENTHUSIASM

Solution 2 (Generalization and Induction)

We can in fact find the probability that any number of randomly distributed numbers on the interval $[0, 1]$ sum to more than $1$ using geometric probability, as shown in the video below.

If we graph the points that satisfy $x + y < 1$, $0 < x, y < 1$, we get the triangle with points $(0, 0)$, $(1, 0)$, and $(0, 1)$. If we graph the points that satisfy $x + y + z < 1$, $0 < x, y, z < 1$, we get the tetrahedron with points $(0, 0, 0)$, $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$.

Of course, the probability of either of these cases happening is simply the area/volume of the points we graphed divided by the total area of the graph, which is always $1$ (this would be much simpler than my calculus proof above).

Thus, we can now solve for the probability that the sum is less than one for $n$ numbers using induction.

$\textbf{Claim:}$ The probability that the sum is less than one is $\frac{1}{n!}$.

$\textbf{Base Case:}$ For just $1$ number, the probability is $1$.

$\textbf{Induction step:}$ Suppose that the probability for $n$ numbers is $\frac{1}{n!}$. We will prove that the probability for $n+1$ numbers is $\frac{1}{(n+1)!}$. To prove this, we consider that the area of an $n+1$-dimensional tetrahedron is simply the area/volume of the base times the height divided by $n+1$.

Of course, the area of the base is $\frac{1}{n!}$, and the height is $1$, and thus, we obtain $\frac{1}{n! \cdot (n+1)} = \frac{1}{(n+1)!}$ as our volume (this may be hard to visualize for higher dimensions). The induction step is complete.

The probability of the sum being less than $1$ is $\frac{1}{n!}$, and the probability of the sum being more than $1$ is $\frac{n!-1}{n!}$. This trivializes the problem. The answer is \[\frac{1}{2} \cdot \frac{2! - 1}{2!} + \frac{1}{2} \cdot \frac{3! - 1}{3!} = \boxed{\textbf{(C) }\frac{2}{3}}.\]

~mathboy100

Solution 3 (Observations)

There are two cases: Amelia takes two steps or three steps.

The former case has a probability of $\frac{1}{2}$, as stated in Solution 1, and thus the latter also has a probability of $\frac{1}{2}$.

The probability that Amelia passes $1$ after two steps is also $\frac{1}{2}$, as it is symmetric to the probability above.

Thus, if the probability that Amelia passes $1$ after three steps is $x$, our total probability is $\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot x$. We know that $0 < x < 1$, and it is relatively obvious that $x > \frac{1}{2}$ (because the probability that $x > \frac{3}{2}$ is $\frac{1}{2}$). This means that our total probability is between $\frac{1}{2}$ and $\frac{3}{4}$, non-inclusive, so the only answer choice that fits is $\boxed{\textbf{(C) }\frac{2}{3}}$.

~mathboy100

Remark (Calculus)

It is not immediately clear why three random numbers between $0$ and $1$ have a probability of $\frac{5}{6}$ of summing to more than $1$. Here is a proof:

Let us start by finding the probability that two random numbers between $0$ and $1$ have a sum of more than $x$, where $0 \leq x \leq 1$.

Suppose that our two numbers are $y$ and $z$. Then, the probability that $y > x$ (which means that $y + z > x$) is $1 - x$, and the probability that $y < x$ is $x$.

If $y < x$, the probability that $y + z > x$ is $1 - x + y$. This is because the probability that $y + z < x$ is equal to the probability that $z < x - y$, which is $x - y$, so our total probability is $1 - (x - y) = 1 - x + y$.

Let us now find the average of the probability that $y + z > x$ when $y < x$. Since $y$ is a random number between $0$ and $x$, its average is $\frac{x}{2}$. Thus, our average is $1 - x + \frac{1}{2} = 1 - \frac{x}{2}$.

Hence, our total probability is equal to \[1(1-x) + \left(1 - \frac{x}{2}\right)(x) = 1 - \frac{1}{2}x^2.\] Now, let us find the probability that three numbers uniformly distributed between $0$ and $1$ sum to more than $1$.

Let our three numbers be $a$, $b$, and $c$. Then, the probability that $a + b + c > 1$ is equal to the probability that $b + c$ is greater than $1 - a$, which is equal to $1 - \frac{1}{2}(1 - a)^2$.

To find the total probability, we must average over all values of $a$. This average is simply equal to the area under the curve $1 - \frac{1}{2}(1-x)^2$ from $0$ to $1$, all divided by $1$. We can compute this value using integrals: \begin{align*} \frac{\int_0^1 \! 1 - \frac{1}{2}(1 - x)^2 \mathrm{d}x}{1} &= \int_0^1 \! 1 - \frac{1}{2}(1 - x)^2 \mathrm{d}x \\ &= 1 - \frac{1}{2}\int_0^1 \! (1 - x)^2 \mathrm{d}x \\ &= 1 - \frac{1}{2}\int_0^1 \! x^2 \mathrm{d}x \\ &= 1 - \frac{1}{2}\left(\frac{1}{3}\right) \\ &= \frac{5}{6}. \end{align*} For those who don't know calculus, $\int_m^n \! f(x) \mathrm{d}x$ is the area under the curve $f(x)$ from $m$ to $n$.

~mathboy100

Remark (Rigorous Calculus)

In the language of probability, we have three random variables $X$, $Y$, $Z$, each independent and uniformly distributed over the interval $(0, 1)$. We are interested in the probability that $X+Y+Z>1$, or $P(X+Y+Z>1)$.

It follows from the definition of the joint probability density function $f(x, y, z)$ that (we assume that $0<x, y, z<1$): \begin{align} P(X+Y+Z>1)=\iiint_{x+y+z>1} f(x,y,z) \,dx\,dy\,dz. \end{align} Since $X, Y, Z$ are independent, we have $f(x, y, z)=f_X(x)*f_Y(y)*f_Z(z)$, where $f_X(x), f_Y(y), f_Z(z)$ represent the probability densities of $X, Y, Z$ respectively. Recall that $X, Y, Z$ are uniformly distributed over the interval $(0, 1)$. Hence, $f_X(x)=f_Y(y)=f_Z(z)=1$.

It remains to evaluate the following triple integral \begin{align} \iiint_{x+y+z>1} \,dx\,dy\,dz=\frac{5}{6}. \end{align} Therefore, $P(X+Y+Z>1)=\frac{5}{6}$.

~tsun26

Video Solution by OmegaLearn Using Geometric Probability

https://youtu.be/-AqhcVX8mTw

~ pi_is_3.14

Video Solution

https://youtu.be/WsA94SmsF5o

~ThePuzzlr

https://youtu.be/qOxnx_c9kVo

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

Video Solution by The Power of Logic (#22 and #23)

https://youtu.be/rZaJSTbs7jY

Video Solution by Interstigation

https://youtu.be/KRkNnlszdEg

~Interstigation

See Also

2022 AMC 10B (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
2022 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

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