2004 AMC 12A Problems/Problem 20
Problem
Select numbers and between and independently and at random, and let be their sum. Let and be the results when and , respectively, are rounded to the nearest integer. What is the probability that ?
Solution
Solution 1
- . The probability that and is . Notice that the sum ranges from to with a symmetric distribution across , and we want . Thus the chance is .
- . The probability that and is , but now , which makes automatically. Hence the chance is .
- . This is the same as the previous case.
- . We recognize that this is equivalent to the first case.
Our answer is .
Solution 2
Use areas to deal with this continuous probability problem. Set up a unit square with values of on x-axis and on y-axis.
If then this will work because . Similarly if then this will work because in order for this to happen, and are each greater than making , and . Each of these triangles in the unit square has area of 1/8.
The only case left is when . Then each of and must be 1 and 0, in any order. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square.
Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is .
Solution 3 (Alcumus)
The conditions under which are as follows.
(i) If , then .
(ii) If and , then and .
(iii) If and , then and .
(iv) If , then and .
These conditions correspond to the shaded regions of the graph shown. The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is .
See Also
2004 AMC 12A (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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