Difference between revisions of "2017 AMC 12B Problems/Problem 20"
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==Problem 20== | ==Problem 20== | ||
| − | Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor | + | Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>? |
<math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math> | <math>\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{1}{4}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{1}{2}</math> | ||
Revision as of 22:49, 14 September 2020
Problem 20
Real numbers 
and 
are chosen independently and uniformly at random from the interval 
. What is the probability that 
?
Solution
First let us take the case that 
. In this case, both and lie in the interval 
. The probability of this is 
. Similarly, in the case that 
, and lie in the interval 
, and the probability is 
. It is easy to see that the probabilities for 
for 
are the infinite geometric series that starts at 
and with common ratio . Using the formula for the sum of an infinite geometric series, we get that the probability is 
.
Solution by: vedadehhc
See Also
| 2017 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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