Difference between revisions of "2022 AMC 10B Problems/Problem 23"
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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> | </cmath> |
Revision as of 14:12, 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
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)