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

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==Problem==
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#redirect [[2022 AMC 12B Problems/Problem 22]]
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>?
 
 
 
<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>
 
==Solution==
 
 
 
We use the following lemma to solve this problem.
 
 
 
---------------------------------------
 
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>,
 
<cmath>
 
\[
 
\Bbb P \left( y_1 + y_2 \leq 1 \right) = \frac{1}{2} .
 
\]
 
</cmath>
 
 
 
For <math>n = 3</math>,
 
<cmath>
 
\[
 
\Bbb P \left( y_1 + y_2 + y_3 \leq 1 \right) = \frac{1}{6} .
 
\]
 
</cmath>
 
---------------------------------------
 
 
 
Now, we solve this problem.
 
 
 
We denote by <math>\tau</math> the last step Amelia moves. Thus, <math>\tau \in \left\{ 2, 3 \right\}</math>.
 
We have
 
 
 
<cmath>
 
\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*}
 
</cmath>
 
 
 
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.
 
 
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
 
 
==Solution 2 (Elimination)==
 
There is a <math>0</math> probability that Amelia is past <math>1</math> after <math>1</math> turn, so Amelia can only pass <math>1</math> after <math>2</math> turns or <math>3</math> turns. The probability of finishing in <math>2</math> turns is <math>\frac{1}{2}</math> (due to the fact that the probability of getting <math>x</math> is the same as the probability of getting <math>2 - x</math>), and thus the probability of finishing in <math>3</math> turns is also <math>\frac{1}{2}</math>.
 
 
 
It is also clear that the probability of Amelia being past <math>1</math> in <math>2</math> turns is equal to <math>\frac{1}{2}</math>.
 
 
 
Therefore, if <math>x</math> is the probability that Amelia finishes if she takes three turns, our final probability is <math>\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot x = \frac{1}{4} + \frac{1}{2} \cdot x</math>.
 
 
 
<math>x</math> must be a number between <math>0</math> and <math>1</math> (non-inclusive), and it is clearly greater than <math>\frac{1}{2}</math>, because the probability of getting more than <math>\frac{3}{2}</math> in <math>3</math> turns is <math>\frac{1}{2}</math>. Thus, the answer must be between <math>\frac{1}{2}</math> and <math>\frac{3}{4}</math>, non-inclusive, so the only answer that makes sense is <math>\fbox{C}</math>.
 
 
 
~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)
 

Revision as of 13:10, 18 November 2022