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Difference between revisions of "2022 AMC 12B Problems/Problem 22"

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<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>
 
<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>
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==Solution==
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Obviously the chance of Amelia stopping after only <math>1</math> step is <math>0</math>.
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When Amelia takes <math>2</math> steps, then the sum of the time taken during the steps is greater than <math>1</math> minute. Let the time taken be <math>x</math> and <math>y</math> respectively, then we need <math>x+y>1</math> for <math>0<x<1, 0<y<1</math>, which has a chance of <math>\frac{1}{2}</math>. Let the lengths of steps be <math>a</math> and <math>b</math> respectively, then we need <math>a+b>1</math> for <math>0<a<1, 0<b<1</math>, which has a chance of <math>\frac{1}{2}</math>. Thus the total chance for this case is <math>\frac{1}{4}</math>.
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When Amelia takes <math>3</math> steps, then by complementary counting the chance of taking <math>3</math> steps is <math>1-\frac{1}{2}=\frac{1}{2}</math>. Let the lengths of steps be <math>a</math>, <math>b</math> and <math>c</math> respectively, then we need <math>a+b+c>1</math> for <math>0<a<1, 0<b<1, 0<c<1</math>, which has a chance of <math>\frac{5}{6}</math>. Thus the total chance for this case is <math>\frac{5}{12}</math>.
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Thus the answer is <math>\frac{1}{4}+\frac{5}{12}=\frac{2}{3}</math>.

Revision as of 07:54, 18 November 2022

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

Obviously the chance of Amelia stopping after only $1$ step is $0$.

When Amelia takes $2$ steps, then the sum of the time taken during the steps is greater than $1$ minute. Let the time taken be $x$ and $y$ respectively, then we need $x+y>1$ for $0<x<1, 0<y<1$, which has a chance of $\frac{1}{2}$. Let the lengths of steps be $a$ and $b$ respectively, then we need $a+b>1$ for $0<a<1, 0<b<1$, which has a chance of $\frac{1}{2}$. Thus the total chance for this case is $\frac{1}{4}$.

When Amelia takes $3$ steps, then by complementary counting the chance of taking $3$ steps is $1-\frac{1}{2}=\frac{1}{2}$. Let the lengths of steps be $a$, $b$ and $c$ respectively, then we need $a+b+c>1$ for $0<a<1, 0<b<1, 0<c<1$, which has a chance of $\frac{5}{6}$. Thus the total chance for this case is $\frac{5}{12}$.

Thus the answer is $\frac{1}{4}+\frac{5}{12}=\frac{2}{3}$.

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