2022 AMC 12B Problems/Problem 22

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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2022 AMC 12B Problems/Problem 22

Problem

Solution