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

Revision as of 02:20, 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

We use the following lemma to solve this problem.

Let $y_1, y_2, \cdots, y_n$ be independent random variables that are uniformly distributed on $(0,1)$ . Then for $n = 2$ , $\Bbb P \left( y_1 + y_2 \leq 1 \right) = \frac{1}{2} .$

For $n = 3$ , $\Bbb P \left( y_1 + y_2 + y_3 \leq 1 \right) = \frac{1}{6} .$

Now, we solve this problem.

We denote by $\tau$ the last step Amelia moves. Thus, $\tau \in \left\{ 2, 3 \right\}$ . We have

$\begin{align*} P \left( \sum_{n=1}^\tau x_n > 1 \right) & = P \left( x_1 + x_2 > 1 | t_1 + t_2 > 1 \right) P \left( t_1 + t_2 > 1 \right) \\ & \hspace{1cm} + P \left( x_1 + x_2 + x_3 > 1 | t_1 + t_2 \leq 1 \right) P \left( t_1 + t_2 \leq 1 \right) \\ & = P \left( x_1 + x_2 > 1 \right) P \left( t_1 + t_2 > 1 \right) + P \left( x_1 + x_2 + x_3 > 1 \right) P \left( t_1 + t_2 \leq 1 \right) \\ & = \left( 1 - \frac{1}{2} \right)\left( 1 - \frac{1}{2} \right) + \left( 1 - \frac{1}{6} \right) \frac{1}{2} \\ & = \boxed{\textbf{(C) } \frac{2}{3}} , \end{align*}$

where the second equation follows from the property that $\left\{ x_n \right\}$ and $\left\{ t_n \right\}$ are independent sequences, the third equality follows from the lemma above.

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

Solution 2 (Elimination)

There is a $0$ probability that Amelia is past $1$ after $1$ turn, so Amelia can only pass $1$ after $2$ turns or $3$ turns. The probability of finishing in $2$ turns is $\frac{1}{2}$ (due to the fact that the probability of getting $x$ is the same as the probability of getting $2 - x$ ), and thus the probability of finishing in $3$ turns is also $\frac{1}{2}$ .

It is also clear that the probability of Amelia being past $1$ in $2$ turns is equal to $\frac{1}{2}$ .

Therefore, if $x$ is the probability that Amelia finishes if she takes three turns, our final probability is $\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot x = \frac{1}{4} + \frac{1}{2} \cdot x$ .

$x$ must be a number between $0$ and $1$ (non-inclusive), and it is clearly greater than $\frac{1}{2}$ , because the probability of getting more than $\frac{3}{2}$ in $3$ turns is $\frac{1}{2}$ . Thus, the answer must be between $\frac{1}{2}$ and $\frac{3}{4}$ , non-inclusive, so the only answer that makes sense is $\fbox{C}$ .

~mathboy100

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)

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