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

Revision as of 22:39, 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 1

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

Solution 3

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 $\frac{1}{4}+\frac{5}{12}=\frac{2}{3}$ .

Solution 4 (Clever)

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

The former case has a probability of $\frac{1}{2}$ , as stated above, 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 $x > 1$ (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

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)

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 12 Problems and Solutions

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

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