Difference between revisions of "2022 AMC 10B Problems/Problem 23"
(→Solution) |
(→Solution) |
||
| Line 37: | Line 37: | ||
& = \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 
,
![\[ \Bbb P \left( y_1 + y_2 \leq 1 \right) = \frac{1}{2} . \]](http://latex.artofproblemsolving.com/9/c/a/9ca0d2902c5eb8db92cddab19e3ca4e92b982504.png)
For 
,
![\[ \Bbb P \left( y_1 + y_2 + y_3 \leq 1 \right) = \frac{1}{6} . \]](http://latex.artofproblemsolving.com/f/7/e/f7e0ecac9af7c72deadc3b60deae91f69748cb18.png)
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)
