Difference between revisions of "1987 AHSME Problems/Problem 26"
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\textbf{(D)}\ \frac{3}{5}\qquad | \textbf{(D)}\ \frac{3}{5}\qquad | ||
\textbf{(E)}\ \frac{3}{4} </math> | \textbf{(E)}\ \frac{3}{4} </math> | ||
| + | |||
| + | == Solution == | ||
| + | The two parts may round to <math>0</math> and <math>2</math>; <math>1</math> and <math>2</math>; <math>1</math> and <math>1</math>; <math>2</math> and <math>1</math>; or <math>2</math> and <math>0</math>. By considering the possible ranges for each case, it is easy to see that each case is equally likely (they divide the interval from <math>0</math> to <math>2.5</math>, in which one of the parts is found, into five equal ranges of <math>0</math> to <math>0.5</math>, <math>0.5</math> to <math>1</math>, <math>1</math> to <math>1.5</math>, <math>1.5</math> to <math>2</math>, and <math>2</math> to <math>2.5</math>). As exactly two of the five cases give a sum of <math>1 + 2 = 3</math>, the answer is <math>\frac{2}{5}</math>, which is answer <math>\boxed{B}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 13:01, 31 March 2018
Problem
The amount 
is split into two nonnegative real numbers uniformly at random,
for instance, into 
and 
, or into 
and 
.
Then each number is rounded to its nearest integer, for instance, 
and 
in the first case above,
and 
in the second. What is the probability that the two integers sum to 
?
Solution
The two parts may round to and ; and ; and ; and ; or and . By considering the possible ranges for each case, it is easy to see that each case is equally likely (they divide the interval from to , in which one of the parts is found, into five equal ranges of to 
, to , to 
, to , and to ). As exactly two of the five cases give a sum of 
, the answer is 
, which is answer 
.
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
