Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Remark)
(→‎Remark)
Line 97: Line 97:
 
<cmath> = 1 - \frac{1}{2}\int_0^1 \! (1 - x)^2 \mathrm{d}x</cmath>
 
<cmath> = 1 - \frac{1}{2}\int_0^1 \! (1 - x)^2 \mathrm{d}x</cmath>
 
<cmath> = 1 - \frac{1}{2}\int_0^1 \! x^2 \mathrm{d}x</cmath>
 
<cmath> = 1 - \frac{1}{2}\int_0^1 \! x^2 \mathrm{d}x</cmath>
−
<cmath> = 1 - \frac{1}{2}\frac{1}{3}</cmath>
+
<cmath> = 1 - \frac{1}{2}(\frac{1}{3})</cmath>
 
<cmath> = \boxed{\frac{5}{6}}.</cmath>
 
<cmath> = \boxed{\frac{5}{6}}.</cmath>
  

Revision as of 22:11, 25 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 $n = 3$, \[ \Bbb P \left( y_1 + y_2 + y_3 \leq 1 \right) = \frac{1}{6} . \textrm{(Check remark for proof)} \]

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 (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


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 $\frac{5}{6}$ (Check remark for proof). Thus the total chance for this case is $\frac{5}{12}$.

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

Remark

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 $1$, its average is $\frac{1}{2}$. Thus, our average is $1 - x + \frac{1}{2} = \frac{1}{2} - x$.

Finally, our total probability is equal to

\[1(1-x) + (\frac{1}{2} - x)(x)\] \[= 1 - \frac{1}{2}x^2.\]

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: (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 $n$)

\[\frac{\int_0^1 \! 1 - \frac{1}{2}(1 - x)^2 \mathrm{d}x}{1}\] \[= \int_0^1 \! 1 - \frac{1}{2}(1 - x)^2 \mathrm{d}x\] \[= 1 - \frac{1}{2}\int_0^1 \! (1 - x)^2 \mathrm{d}x\] \[= 1 - \frac{1}{2}\int_0^1 \! x^2 \mathrm{d}x\] \[= 1 - \frac{1}{2}(\frac{1}{3})\] \[= \boxed{\frac{5}{6}}.\]

~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)

See Also

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12B_Problems/Problem_22&oldid=183032"