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

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#redirect [[2022 AMC 12B Problems/Problem 22]]
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==Problem==
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Ant Amelia starts on the number line at <math>0</math> and crawls in the following manner. For <math>n=1,2,3,</math> Amelia chooses a time duration <math>t_n</math> and an increment <math>x_n</math> independently and uniformly at random from the interval <math>(0,1).</math> During the <math>n</math>th step of the process, Amelia moves <math>x_n</math> units in the positive direction, using up <math>t_n</math> minutes. If the total elapsed time has exceeded <math>1</math> minute during the <math>n</math>th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most <math>3</math> steps in all. What is the probability that Amelia’s position when she stops will be greater than <math>1</math>?
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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>
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==Solution 1 (Geometric Probability)==
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==Solution 2 (Geometric Probability and Calculus)==
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We use the following lemma to solve this problem.
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Let <math>y_1, y_2, \cdots, y_n</math> be independent random variables that are uniformly distributed on <math>(0,1)</math>. Then for <math>n = 2</math>,
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<cmath>
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\[
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\Bbb P \left( y_1 + y_2 \leq 1 \right) = \frac{1}{2} .
 +
\]
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</cmath>
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For <math>n = 3</math>,
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<cmath>
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\[
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\Bbb P \left( y_1 + y_2 + y_3 \leq 1 \right) = \frac{1}{6}.
 +
\]
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</cmath>
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Check the <b>Remark</b> section for proof.
 +
 
 +
Now, we solve this problem.
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We denote by <math>\tau</math> the last step Amelia moves. Thus, <math>\tau \in \left\{ 2, 3 \right\}</math>.
 +
We have
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<cmath>
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\begin{align*}
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P \left( \sum_{n=1}^\tau x_n > 1 \right)
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& = P \left( x_1 + x_2 > 1 | t_1 + t_2 > 1 \right)
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P \left( t_1 + t_2 > 1 \right) \
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& \hspace{1cm} + P \left( x_1 + x_2 + x_3 > 1 | t_1 + t_2 \leq 1 \right)
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P \left( t_1 + t_2 \leq 1 \right) \
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& = P \left( x_1 + x_2 > 1 \right)
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P \left( t_1 + t_2 > 1 \right)
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+ P \left( x_1 + x_2 + x_3 > 1 \right)
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P \left( t_1 + t_2 \leq 1 \right) \
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& = \left( 1 - \frac{1}{2} \right)\left( 1 - \frac{1}{2} \right)
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+ \left( 1 - \frac{1}{6} \right) \frac{1}{2} \
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& = \boxed{\textbf{(C) }\frac{2}{3}},
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\end{align*}
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</cmath>
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where the second equation follows from the property that <math>\left\{ x_n \right\}</math> and <math>\left\{ t_n \right\}</math> are independent sequences, the third equality follows from the lemma above.
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 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
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==Solution 3 (Casework)==
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There are two cases: Amelia takes two steps or three steps.
 +
 
 +
The former case has a probability of <math>\frac{1}{2}</math>, as stated in Solution 2, and thus the latter also has a probability of <math>\frac{1}{2}</math>.
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 +
The probability that Amelia passes <math>1</math> after two steps is also <math>\frac{1}{2}</math>, as it is symmetric to the probability above.
 +
 
 +
Thus, if the probability that Amelia passes <math>1</math> after three steps is <math>x</math>, our total probability is <math>\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot x</math>. We know that <math>0 < x < 1</math>, and it is relatively obvious that <math>x > \frac{1}{2}</math> (because the probability that <math>x > \frac{3}{2}</math> is <math>\frac{1}{2}</math>). This means that our total probability is between <math>\frac{1}{2}</math> and <math>\frac{3}{4}</math>, non-inclusive, so the only answer choice that fits is <math>\boxed{\textbf{(C) }\frac{2}{3}}</math>.
 +
 
 +
~mathboy100
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==Solution 4 (Generalization and Induction)==
 +
 
 +
We can in fact find the probability that any number of randomly distributed numbers on the interval <math>[0, 1]</math> sum to more than <math>1</math> using geometric probability, as shown in the video below.
 +
 
 +
If we graph the points that satisfy <math>x + y < 1</math>, <math>0 < x, y < 1</math>, we get the triangle with points <math>(0, 1)</math>, <math>(0, 0)</math>, and <math>(1, 0)</math>. If we graph the points that satisfy <math>x + y + z < 1</math>, <math>0 < x, y, z < 1</math>, we get the tetrahedron with points <math>(0, 1, 0)</math>, <math>(0, 0, 0)</math>, <math>0, 0, 1</math>, and <math>(1, 0, 0)</math>.
 +
 
 +
Of course, the probability of either of these cases happening is simply the area/volume of the points we graphed divided by the total area of the graph, which is always <math>1</math> (this would be much simpler than my calculus proof above).
 +
 
 +
Thus, we can now solve for the probability that the sum is less than one for <math>n</math> numbers using induction.
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<math>\textbf{Claim:}</math> The probability that the sum is less than one is <math>\frac{1}{n!}</math>.
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<math>\textbf{Base Case:}</math> For just <math>1</math> number, the probability is <math>1</math>.
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<math>\textbf{Induction step:}</math> Suppose that the probability for <math>n</math> numbers is <math>\frac{1}{n!}</math>. We will prove that the probability for <math>n+1</math> numbers is <math>\frac{1}{(n+1)!}</math>. To prove this, we consider that the area of an <math>n+1</math>-dimensional tetrahedron is simply the area/volume of the base times the height divided by <math>n+1</math>.
 +
 
 +
Of course, the area of the base is <math>\frac{1}{n!}</math>, and the height is <math>1</math>, and thus, we obtain <math>\frac{1}{n! \cdot (n+1)} = \frac{1}{(n+1)!}</math> as our volume (this may be hard to visualize for higher dimensions). The induction step is complete.
 +
 
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The probability of the sum being less than <math>1</math> is <math>\frac{1}{n!}</math>, and the probability of the sum being more than <math>1</math> is <math>\frac{n!-1}{n!}</math>. This trivializes the problem. The answer is <cmath>\frac{1}{2} \cdot \frac{2! - 1}{2!} + \frac{1}{2} \cdot \frac{3! - 1}{3!} = \boxed{\textbf{(C) }\frac{2}{3}}.</cmath>
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 +
~mathboy100
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==Remark==
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It is not immediately clear why three random numbers between <math>0</math> and <math>1</math> have a probability of <math>\frac{5}{6}</math> of summing to more than <math>1</math>. Here is a proof:
 +
 
 +
Let us start by finding the probability that two random numbers between <math>0</math> and <math>1</math> have a sum of more than <math>x</math>, where <math>0 \leq x \leq 1</math>.
 +
 
 +
Suppose that our two numbers are <math>y</math> and <math>z</math>. Then, the probability that <math>y > x</math> (which means that <math>y + z > x</math>) is <math>1 - x</math>, and the probability that <math>y < x</math> is <math>x</math>.
 +
 
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If <math>y < x</math>, the probability that <math>y + z > x</math> is <math>1 - x + y</math>. This is because the probability that <math>y + z < x</math> is equal to the probability that <math>z < x - y</math>, which is <math>x - y</math>, so our total probability is <math>1 - (x - y) = 1 - x + y</math>.
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 +
Let us now find the average of the probability that <math>y + z > x</math> when <math>y < x</math>. Since <math>y</math> is a random number between <math>0</math> and <math>x</math>, its average is <math>\frac{x}{2}</math>. Thus, our average is <math>1 - x + \frac{1}{2} = 1 - \frac{x}{2}</math>.
 +
 
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Hence, our total probability is equal to
 +
<cmath>1(1-x) + \left(1 - \frac{x}{2}\right)(x) = 1 - \frac{1}{2}x^2.</cmath>
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Now, let us find the probability that three numbers uniformly distributed between <math>0</math> and <math>1</math> sum to more than <math>1</math>.
 +
 
 +
Let our three numbers be <math>a</math>, <math>b</math>, and <math>c</math>. Then, the probability that <math>a + b + c > 1</math> is equal to the probability that <math>b + c</math> is greater than <math>1 - a</math>, which is equal to <math>1 - \frac{1}{2}(1 - a)^2</math>.
 +
 
 +
To find the total probability, we must average over all values of <math>a</math>. This average is simply equal to the area under the curve <math>1 - \frac{1}{2}(1-x)^2</math> from <math>0</math> to <math>1</math>, all divided by <math>1</math>. We can compute this value using integrals: (for those who don't know calculus, <math>\int_m^n \! f(x) \mathrm{d}x</math> is the area under the curve <math>f(x)</math> from <math>m</math> to <math>n</math>)
 +
<cmath>\begin{align*}
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\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 \
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&= 1 - \frac{1}{2}\int_0^1 \! (1 - x)^2 \mathrm{d}x \
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&= 1 - \frac{1}{2}\int_0^1 \! x^2 \mathrm{d}x \
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&= 1 - \frac{1}{2}\left(\frac{1}{3}\right) \
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&= \boxed{\frac{5}{6}}.
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\end{align*}</cmath>
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~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==
 +
{{AMC10 box|year=2022|ab=B|num-b=22|num-a=24}}
 +
{{AMC12 box|year=2022|ab=B|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Revision as of 11:58, 4 December 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 (Geometric Probability)

Solution 2 (Geometric Probability and Calculus)

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 $n = 3$, \[ \Bbb P \left( y_1 + y_2 + y_3 \leq 1 \right) = \frac{1}{6}. \] Check the Remark section 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 3 (Casework)

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

The former case has a probability of $\frac{1}{2}$, as stated in Solution 2, 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 $x > \frac{1}{2}$ (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 4 (Generalization and Induction)

We can in fact find the probability that any number of randomly distributed numbers on the interval $[0, 1]$ sum to more than $1$ using geometric probability, as shown in the video below.

If we graph the points that satisfy $x + y < 1$, $0 < x, y < 1$, we get the triangle with points $(0, 1)$, $(0, 0)$, and $(1, 0)$. If we graph the points that satisfy $x + y + z < 1$, $0 < x, y, z < 1$, we get the tetrahedron with points $(0, 1, 0)$, $(0, 0, 0)$, $0, 0, 1$, and $(1, 0, 0)$.

Of course, the probability of either of these cases happening is simply the area/volume of the points we graphed divided by the total area of the graph, which is always $1$ (this would be much simpler than my calculus proof above).

Thus, we can now solve for the probability that the sum is less than one for $n$ numbers using induction.

$\textbf{Claim:}$ The probability that the sum is less than one is $\frac{1}{n!}$.

$\textbf{Base Case:}$ For just $1$ number, the probability is $1$.

$\textbf{Induction step:}$ Suppose that the probability for $n$ numbers is $\frac{1}{n!}$. We will prove that the probability for $n+1$ numbers is $\frac{1}{(n+1)!}$. To prove this, we consider that the area of an $n+1$-dimensional tetrahedron is simply the area/volume of the base times the height divided by $n+1$.

Of course, the area of the base is $\frac{1}{n!}$, and the height is $1$, and thus, we obtain $\frac{1}{n! \cdot (n+1)} = \frac{1}{(n+1)!}$ as our volume (this may be hard to visualize for higher dimensions). The induction step is complete.

The probability of the sum being less than $1$ is $\frac{1}{n!}$, and the probability of the sum being more than $1$ is $\frac{n!-1}{n!}$. This trivializes the problem. The answer is \[\frac{1}{2} \cdot \frac{2! - 1}{2!} + \frac{1}{2} \cdot \frac{3! - 1}{3!} = \boxed{\textbf{(C) }\frac{2}{3}}.\]

~mathboy100

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

Hence, our total probability is equal to \[1(1-x) + \left(1 - \frac{x}{2}\right)(x) = 1 - \frac{1}{2}x^2.\] 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 $n$) \begin{align*} \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}\left(\frac{1}{3}\right) \\ &= \boxed{\frac{5}{6}}. \end{align*} ~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 10B (ProblemsAnswer KeyResources)
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
2022 AMC 12B (ProblemsAnswer KeyResources)
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

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