Difference between revisions of "2022 AMC 10B Problems/Problem 23"
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We have | We have | ||
− | < | + | <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) | ||
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& = \left( 1 - \frac{1}{2} \right)\left( 1 - \frac{1}{2} \right) | & = \left( 1 - \frac{1}{2} \right)\left( 1 - \frac{1}{2} \right) | ||
+ \left( 1 - \frac{1}{6} \right) \frac{1}{2} \ | + \left( 1 - \frac{1}{6} \right) \frac{1}{2} \ | ||
− | & = \boxed{\textbf{(C) | + | & = \boxed{\textbf{(C) \frac{2}{3}}} , |
\end{align*} | \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. | 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. |
Revision as of 14:11, 17 November 2022
Solution
We use the following lemma to solve this problem.
Let be independent random variables that are uniformly distributed on . Then for ,
For ,
Now, we solve this problem.
We denote by the last step Amelia moves. Thus, . 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*} (Error compiling LaTeX. Unknown error_msg)
where the second equation follows from the property that and are independent sequences, the third equality follows from the lemma above.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)