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

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We have
 
We have
  
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<math></math>
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<cmath>
 
\begin{align*}
 
\begin{align*}
 
P \left( \sum_{n=1}^\tau x_n > 1 \right)
 
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( x_1 + x_2 > 1 | t_1 + t_2 > 1 \right)
 
  P \left( t_1 + t_2 > 1 \right) \\
 
  P \left( t_1 + t_2 > 1 \right) \\
 +
\end{align*}
 +
</cmath>
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 +
<math></math>
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\begin{align*}
 
& \hspace{1cm} + P \left( x_1 + x_2 + x_3 > 1 | t_1 + t_2 \leq 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( t_1 + t_2 \leq 1 \right) \\

Revision as of 14:09, 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) \\ \end{align*}

$$ (Error compiling LaTeX. Unknown error_msg) \begin{align*} & \hspace{1cm} + P \left( x_1 + x_2 + x_3 > 1 | t_1 + t_2 \leq 1 \right) P \left( t_1 + t_2 \leq 1 \right) \\ & = P \left( x_1 + x_2 > 1 \right) P \left( t_1 + t_2 > 1 \right) + P \left( x_1 + x_2 + x_3 > 1 \right) P \left( t_1 + t_2 \leq 1 \right) \\ & = \left( 1 - \frac{1}{2} \right)\left( 1 - \frac{1}{2} \right) + \left( 1 - \frac{1}{6} \right) \frac{1}{2} \\ & = \boxed{\textbf{(C) $\frac{2}{3}$}} , \end{align*} $$ (Error compiling LaTeX. Unknown error_msg)

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)

Video Solution

https://youtu.be/qOxnx_c9kVo

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

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