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Difference between revisions of "2021 April MIMC 10 Problems/Problem 8"

(Created page with "In the morning, Mr.Gavin always uses his alarm to wake him up. The alarm is special. It always rings in a cycle of ten rings. The first ring lasts <math>1</math> second, and e...")
 
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==Solution==
 
==Solution==
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To be Released on April 26th.
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First, we want to find the total length of a cycle. The length of a ten-rings cycle is equal to <math>2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9=2^{10}-1=1023</math>. The probability is therefore <math>\frac{\textrm{length of the tenth ring}}{\textrm{Total time}}=</math><math>\boxed{\textbf{(C)} \frac{512}{1023}}</math>

Latest revision as of 12:35, 26 April 2021

In the morning, Mr.Gavin always uses his alarm to wake him up. The alarm is special. It always rings in a cycle of ten rings. The first ring lasts $1$ second, and each ring after lasts twice the time than the previous ring. Given that Mr.Gavin has an equal probability of waking up at any time, what is the probability that Mr.Gavin wakes up and end the alarm during the tenth ring?

$\textbf{(A)} ~\frac{511}{1023} \qquad\textbf{(B)} ~\frac{1}{2} \qquad\textbf{(C)} ~\frac{512}{1023} \qquad\textbf{(D)} ~\frac{257}{512} \qquad\textbf{(E)} ~\frac{129}{256}$

Solution

First, we want to find the total length of a cycle. The length of a ten-rings cycle is equal to $2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9=2^{10}-1=1023$. The probability is therefore $\frac{\textrm{length of the tenth ring}}{\textrm{Total time}}=$$\boxed{\textbf{(C)} \frac{512}{1023}}$

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