Difference between revisions of "2022 AMC 10B Problems/Problem 23"
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==Video Solution== | ==Video Solution== | ||
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+ | https://youtu.be/WsA94SmsF5o | ||
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+ | ~ThePuzzlr | ||
https://youtu.be/qOxnx_c9kVo | https://youtu.be/qOxnx_c9kVo | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 21:19, 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)
Solution 2 (Elimination)
There is a probability that Amelia is past after turn, so Amelia can only pass after turns or turns. The probability of finishing in turns is (due to the fact that the probability of getting is the same as the probability of getting ), and thus the probability of finishing in turns is also .
It is also clear that the probability of Amelia being past in turns is equal to .
Therefore, if is the probability that Amelia finishes if she takes three turns, our final probability is .
must be a number between and (non-inclusive), and it is clearly greater than , because the probability of getting more than in turns is . Thus, the answer must be between and , non-inclusive, so the only answer that makes sense is .
~mathboy100
Video Solution
~ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)