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

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==Solution 2 (Elimination)==
 
==Solution 2 (Elimination)==
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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>.
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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>, because this probability is exactly the same as the probability of finishing in <math>2</math> turns.  
 
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>, because this probability is exactly the same as the probability of finishing in <math>2</math> turns.  

Revision as of 15:23, 17 November 2022

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 $\frac{1}{2}$, because this probability is exactly the same as the probability of finishing in $2$ turns.

Therefore, if $x$ is the probability that Amelia finishes in 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 $\frac{3}{2}$ is $\frac{1}{2}$. Thus, the answer must be between $\frac{1}{2}$ and $\frac{3}{4}$, non-inclusive, so the answer is $\fbox{C. \frac{1}{3}}$ (Error compiling LaTeX. Unknown error_msg)

Video Solution

https://youtu.be/qOxnx_c9kVo

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

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