Difference between revisions of "2021 April MIMC 10 Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | 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}}=\fbox{\textbf{(C) | + | 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}}=\fbox{\textbf{(C) \frac{512}{1023}}}</math> |
Revision as of 13:35, 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 
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?
Solution
First, we want to find the total length of a cycle. The length of a ten-rings cycle is equal to 
. The probability is therefore $\frac{\textrm{length of the tenth ring}}{\textrm{Total time}}=\fbox{\textbf{(C) \frac{512}{1023}}}$ (Error compiling LaTeX. Unknown error_msg)
